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Question 1: An isosceles trapezium circumscribed over a circle has one of its parallel sides thrice the other. If the perimeter of the trapezium is 16 cms. The area of the trapezium in sq cm is
a) $4\sqrt{3}$
b) 8
c) $8\sqrt{3}$
d) $16\sqrt{3}$
1) Answer (C)
Solution:
Let the length of one of the parallel sides = 2x
Length of other parallel side = 6x
Since it is isosceles trapezium, the non parallel sides are equal
a+2x=a+2y
$\therefore$ x=y
a=2x
Given Perimeter = 16 cms
16x=16
x=1
Let h be the length of the height dropped to the base
$(4x)^{2}$=$h^{2}+(2x)^{2}$
h=$2\sqrt{3}$
Area of the trapezium= $\frac{1}{2}$*height*(Sum of length of the parallel sides)
=$\frac{1}{2}*2\sqrt{3}*(8)$
=$8\sqrt{3}$
Hence C is the correct answer.
Question 2: A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?
2) Answer: 48
Solution:
Let the efficiency of type A pipe be ‘a’ and the efficiency of type B be ‘b’.
In the first case, 10 type A and 45 type B pipes fill the tank in 30 mins.
So, the capacity of the tank = $\dfrac{1}{2}$(10a + 45b)……..(i)
In the second case, 8 type A and 18 type B pipes fill the tank in 1 hour.
So, the capacity of the tank = (8a + 18b)……….(ii)
Equating (i) and (ii), we get
10a + 45b = 16a + 36b
=>6a = 9b
From (ii), capacity of the tank = (8a + 18b) = (8a + 12a) = 20a
In the third case, 7 type A and 27 type B pipes fill the tank.
Net efficiency = (7a + 27b) = (7a + 18a) = 25a
Time taken = $\dfrac{20\text{a}}{25\text{a}}$ hour = 48 minutes.
Hence, 48 is the correct answer.
Question 3: How many pairs (m, n) of positive integers satisfy the equation $m^2 + 105 = n^2$?
3) Answer: 4
Solution:
$n^2-m^2=105$
(n-m)(n+m) = 1*105, 3*35, 5*21, 7*15, 15*7, 21*5, 35*3, 105*1.
n-m=1, n+m=105 ==> n=53, m=52
n-m=3, n+m=35 ==> n=19, m=16
n-m=5, n+m=21 ==> n=13, m=8
n-m=7, n+m=15 ==> n=11, m=4
n-m=15, n+m=7 ==> n=11, m=-4
n-m=21, n+m=5 ==> n=13, m=-8
n-m=35, n+m=3 ==> n=19, m=-16
n-m=105, n+m=1 ==> n=53, m=-52
Since only positive integer values of m and n are required. There are 4 possible solutions.
Question 4: If the rectangular faces of a brick have their diagonals in the ratio $3 : 2 \surd3 : \surd{15}$, then the ratio of the length of the shortest edge of the brick to that of its longest edge is
a) $\sqrt{3} : 2$
b) $1 : \sqrt{3}$
c) $2 : \sqrt{5}$
d) $\sqrt{2} : \sqrt{3}$
4) Answer (B)
Solution:
Assuming the dimensions of the brick are a, b and c and the diagonals are 3, 2 $\surd3$ and $\surd{15}$
Hence, $a^{2\ }+\ b^2$ = $3^2$ ……(1)
$b^{2\ }+\ c^2$ = $(2\sqrt{3})^2$ ……(2)
$c^{2\ }+\ a^2$ = $(\sqrt{15})^2$ ……(3)
Adding the three equations, 2($a^2+b^2+c^2$) = 9+12+15=36
=>$a^2+b^2+c^2$ = 18……(4)
Subtracting (1) from (4), we get $c^2$ = 9 =>c=3
Subtracting (2) from (4), we get $a^2$ = 6 =>a=$\sqrt{6}$
Subtracting (3) from (4), we get $b^2$ = 3 =>b=$\sqrt{3}$
The ratio of the length of the shortest edge of the brick to that of its longest edge is = $\ \frac{\ \sqrt{3}}{3}$ = $1 : \sqrt{3}$
Question 5: One can use three different transports which move at 10, 20, and 30 kmph, respectively to reach from A to B. Amal took each mode of transport for $\frac{1}{3}^{rd}$ of his total journey time, while Bimal took each mode of transport for $\frac{1}{3}^{rd}$ of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to
a) 22
b) 20
c) 19
d) 21
5) Answer (A)
Solution:
Assume the total distance between A and B as d and time taken by Amal = t
Since Amal travelled $\frac{1}{3}^{rd}$ of his total journey time in different speeds
d = $\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$
$\text{Total time taken by Bimal} = \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$
$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$
Hence, the ratio of time taken by Bimal to time taken by Amal = $\frac{\frac{11t}{9}}{t}=\frac{11}{9}$
Therefore, Bimal will exceed Amal’s time by $\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$
Question 6: The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was
a) $18 \pi$
b) $14 \pi$
c) $16 \pi$
d) $12 \pi$
6) Answer (C)
Solution:
Distance covered by A in 1 revolution = 2$\pi\ $*30 = 60$\pi\ $
Distance covered by B in 1 revolution = 2$\pi\ $*40 = 80$\pi\ $
Now, (5000+n)60$\pi\ $ = 80$\pi\ $n
=> 15000= 4n-3n =>n=15000
Then distance travelled by B = 15000*80$\pi\ $ cm = 12$\pi\ $ km
Hence, the speed = $\ \frac{\ 12\pi\times\ 60\ }{45}$ = 16$\pi\ $
Question 7: A trapezium $ABCD$ has side $AD$ parallel to $BC, \angle BAD = 90^\circ, BC = 3$ cm and $AD= 8$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is
7) Answer: 66
Solution:
CD = $\sqrt{\ y^2+25}$
$11+y+\sqrt{y^2+25}=36$
$\sqrt{y^2+25}=25-y$
$y^2+25=25^2+y^2-50y$
2y = 24
y = 12
Area of trapezium = $3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$
Question 8: Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?
8) Answer: 15
Solution:
Train T starts at 3 PM and train S starts at 4 PM.
Let the speed of train T be t.
=> Speed of train S = 0.75t.
When the trains meet, train t would have traveled for one more hour than train S.
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours.
Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t
We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities.
=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25
x = 9 hours.
Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3) = 15 hours.
Therefore, 15 is the correct answer.
Question 9: For any two positive integer ‘a’ and ‘b’, what is the product of all possible values of ‘a’ for which 1/a + 1/b = 2/9 and a<b
a) 24
b) 30
c) 12
d) 48
9) Answer (B)
Solution:
1/a + 1/b = 2/9
Taking LCM, we get (a+b)/ab = 2/9
Cross multiplying, we get 9a+9b = 2ab
9a = 2ab-9b
9a = b(2a-9)
b = 9a/(2a-9). The only solutions for (a,b) are {(5,45) and (6,18)}
Question 10: One third of the buses from City A to City B stop at City C, while the rest go non-stop to City B. One third of the passengers, in the buses stopping at City C, continue to City B, while the rest alight at City C. All the buses have equal capacity and always start full from City A. What proportion of the passengers going to City B from City A travel by a bus stopping at City C?
a) $\frac{1}{7}$
b) $\frac{1}{9}$
c) $\frac{1}{3}$
d) $\frac{7}{9}$
e) $\frac{4}{9}$
10) Answer (A)
Solution:
Let us assume there are 9 buses.
3 of them stop at C and 6 go non-stop
Given, One-third of the passengers, in the buses stopping at City C, continue to City B, while the rest alight at City C
=> Since all buses have equal capacity. we can say 2 will elite at C and 1 will proceed to B.
Hence required proportion = 1/7
Question 11: A cyclist and a bus start from cities P and Q respectively at the same time. Every day the cyclist and the bus cross each other at a road-side Dhaba. One day the cyclist started 42 minutes later than his usual time. As a result, the bus and the cyclist meet at a point which is 5km away from the restaurant. Find out the speed(in km/h) of the bus, if the cyclist cycled at a uniform speed of 10 km/h.
11) Answer: 25
Solution:
The above figure represents the movement of the cyclist and the bus between cities P and Q.
As the cyclist started 42 minutes later than his usual time but the bus started at its usual time. So, when the bus would have reached the Dhaba (their usual meeting point), the cyclist would be 42 minutes away from the Dhaba.
Distance covered by cyclist in 42 minutes = $\frac{42}{60}$hr$\times 10$ km/hr = 7 km
Let, their new meeting point be somewhere between X and Dhaba. The cyclist has reached point X when Bus has reached the Dhaba. The distance between X and Dhaba is 7km. Since, they meet at a point 5 km away from dhaba. So, the cyclist must have cycled for 2 Km which would have taken him 12 minutes (=$\frac{2}{10}\times 60$).
So, the bus would also have travelled 5km in 12 minutes after crossing the dhaba to meet the cyclist.
Speed of bus = $\frac{5}{\frac{12}{60}}$ km/hr = 25 km/hr.
Hence, the correct answer is 25 km/hr.
Question 12: If a and b are integers of opposite signs such that $(a + 3)^{2} : b^{2} = 9 : 1$ and $(a -1)^{2}:(b – 1)^{2} = 4:1$, then the ratio $a^{2} : b^{2}$ is
a) 9:4
b) 81:4
c) 1:4
d) 25:4
12) Answer (D)
Solution:
Since the square root can be positive or negative we will get two cases for each of the equation.
For the first one,
a + 3 = 3b .. i
a + 3 = -3b … ii
For the second one,
a – 1 = 2(b -1) … iii
a – 1 = 2 (1 – b) … iv
we have to solve i and iii, i and iv, ii and iii, ii and iv.
Solving i and iii,
a + 3 = 3b and a = 2b – 1, solving, we get a = 3 and b = 2, which is not what we want.
Solving i and iv
a + 3 = 3b and a = 3 – 2b, solving, we get b = 1.2, which is not possible.
Solving ii and iii
a + 3 = -3b and a = 2b – 1, solving, we get b = 0.4, which is not possible.
Solving ii and iv,
a + 3 = -3b and a = 3 – 2b, solving, we get a = 15 and b = -6 which is what we want.
Thus, $\frac{a^2}{b^2} = \frac{25}{4}$
Question 13: Ravi appeared in 5 exams – English, Maths, Science, Social science and Computer science. The total marks allotted for each subject is different and is a multiple of 50. 40% of the total marks allotted to a subject is considered pass marks. The total marks allotted for the 5 subjects put together is 750. Ravi passed in each subject. The difference between his marks in English and the pass mark in English is 10% of the pass marks. For Maths, this difference is 20% of the pass marks, 30% for Science, 40% for Social science and 50% for computer science. If Ravi obtained 410 marks in total, the difference between the marks obtained by him in Maths and Social Science is
a) 64
b) 56
c) 34
d) Cannot be determined
13) Answer (A)
Solution:
It has been given that the total marks in each subject is different and is a multiple of 50.
Let the total marks in English be 50a, Maths be 50b, Science be 50c, Social science be 50d and Computer science be 50e, where a,b,c,d, and e are different integers.
Total marks allotted for the 5 subjects = 750
=> 50a + 50b + 50c + 50d + 50e = 750
a + b + c + d + e = 15.
The least value that ‘a’ can take is 1.
1 + 2 + 3 + 4 + 5 = 15.
Therefore, the marks allotted to the 5 subjects should be 50 , 100, 150, 200, and 250.
40% of the total marks allotted to a subject is pass marks.
Therefore, the pass marks in the 5 subjects are 20, 40, 60, 80, and 100.
It has been given that Ravi obtained 10% more than the pass marks in English, 20% more than the pass marks on Maths, 30% more than the pass marks in Science, 40% more than the pass marks in Social science and 50% more than the pass marks in computer science.
Had Ravi obtained the pass marks in all the subjects, he would have received 20 + 40 + 60 + 80 + 100 = 300 marks.
We know that Ravi has obtained 410 marks in total or 110/300 = 11/30 more than the pass marks.
The pass marks are in the ratio 1:2:3:4:5. Let the pass marks be x, 2x , 3x, 4x and 5x. Had Ravi obtained the pass marks, he would have obtained 15x. However, Ravi has obtained 15x + 15x*11/30 = 20.5x in total.
We have to multiply 1.1, 1.2, 1.3, 1.4 and 1.5 with x, 2x, 3x, 4x, and 5x such that the sum of the values obtained is 20.5x. Let us find out the least possible and maximum possible values that can be obtained using the combinations.
The least possible value will occur when 1.1 is multiplied with 5x, 1.2 is multiplied with 4x, 1.3 is multiplied with 3x, 1.4 is multiplied with 4x and 1.5 is multiplied with 5x.
In this case, the total marks obtained by Ravi will be 1.1*5x + 1.2*4x + 1.3*3x + 1.4*2x + 1.5*x
= 5.5x + 4.8x + 3.9x + 2.8x + 1.5x = 18.5x.
The maximum possible value will occur when 1.1*x + 1.2*2x + 1.3*3x + 1.4*4x + 1.5*5x = 1.1x + 2.4x + 3.9x + 5.6x + 7.5x = 20.5x.
The given case represents the maximum possible score that could have been scored by Ravi for the given conditions.
Therefore, the English exam must be of 50 marks, Maths exam must be of 100 marks, Science exam must be of 150 marks, Social science exam must be of 200 marks and Computer science exam must be of 250 marks.
Difference between the marks obtained by Ravi in Maths and Social Science = 200*0.4*1.4 – 100*0.4*1.2 = 112 – 48 = 64.
Therefore, option A is the right answer.
Question 14: Brother and sister both appear for an interview. The probability of the selection of brother is $\frac{1}{8}$ while the probability of rejection of sister is $\frac{4}{5}$ What is the probability that only one of them is selected ?
a) $\frac{11}{40}$
b) $\frac{7}{40}$
c) $\frac{5}{13}$
d) $\frac{1}{10}$
14) Answer (A)
Solution:
Probability of brother’s selection = $\frac{1}{8}$ and probability of brother’s rejection = $\frac{7}{8}$
Probability of sister’s selection = $\frac{1}{5}$ and probability of sister’s rejection = $\frac{4}{5}$
Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)
= $(\frac{1}{8})(\frac{4}{5})+(\frac{7}{8})(\frac{1}{5})$
= $\frac{4}{40}+\frac{7}{40}=\frac{11}{40}$
=> Ans – (A)
Question 15: A shop keeper uses a 102 cm scale while buying clothes. He sells clothes using a 98 cm scale. If he marks up the price of the cloth by 10%, what is the maximum discount (approximately) he can offer without incurring a loss?
a) 15.50%
b) 14.50%
c) 10.50%
d) 12.65%
e) 8.50%
15) Answer (D)
Solution:
Let us assume the CP of 100 cm of cloth to be Rs. 100.
The shop keeper will buy 102 cm for Rs. 100.
He marks up the price by 10%. He will sell cloth at Rs. 110/m.
But he is using a 98 cm scale. Therefore, he will sell only 98 cm for Rs. 110.
Let us assume that the shop keeper provides a discount of (100 – x)%. He will sell the cloth at a price of x*110
This must be equal to the cost incurred by the shop keeper for procuring 98 cm of cloth.
The shopkeeper procures 102 cm at Rs. 100. Therefore, he will procure 98 cm at 100*98/102 = Rs. 96.07(approx.).
x*110 = 96.07
=> x = 87.35%
Therefore, the maximum discount that can be offered is around 12.65% (12.5% approximately).
Question 16: If $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$ , then what is the value of $U+3V$ ?
a) $0$
b) $\dfrac{1}{2}$
c) $\dfrac{-1}{4}$
d) $\dfrac{1}{4}$
16) Answer (C)
Solution:
Given that $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$
$\Rightarrow$ $U^{2}+(U-2V-1)(U-2V-1)$= −$4V(U+V)$
$\Rightarrow$ $U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$ = −$4V(U+V)$
$\Rightarrow$ $U^{2}+(U^2-4UV-2U+4V^2+4V+1)$ = −$4V(U+V)$
$\Rightarrow$ $2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$
$\Rightarrow$ $2U^2-2U+8V^2+4V+1=0$
$\Rightarrow$ $2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$
$\Rightarrow$ $2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$
Sum of two square terms is zero i.e. individual square term is equal to zero.
$U-\dfrac{1}{2}$ = 0 and $V+\dfrac{1}{4}$ = 0
U = $\dfrac{1}{2}$ and V = $-\dfrac{1}{4}$
Therefore, $U+3V$ = $\dfrac{1}{2}$+$\dfrac{-1*3}{4}$ = $\dfrac{-1}{4}$. Hence, option C is the correct answer.
Question 17: If $a_1, a_2, ……$ are in A.P., then, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$ is equal to
a) $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$
b) $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_{n – 1}}}$
c) $\frac{n – 1}{\sqrt{a_1} + \sqrt{a_n}}$
d) $\frac{n}{\sqrt{a_1} – \sqrt{a_{n + 1}}}$
17) Answer (A)
Solution:
We have, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$
Now, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$ = $\frac{\sqrt{a_2} – \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} – \sqrt{a_1})}$ (Multiplying numerator and denominator by $\sqrt{a_2} – \sqrt{a_1}$)
= $\frac{\sqrt{a_2} – \sqrt{a_1}}{({a_2} – {a_1}}$
=$\frac{\sqrt{a_2} – \sqrt{a_1}}{d}$ (where d is the common difference)
Similarly, $ \frac{1}{\sqrt{a_2} + \sqrt{a_3}}$ = $\frac{\sqrt{a_3} – \sqrt{a_2}}{d}$ and so on.
Then the expression $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ……. + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$
can be written as $\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+……………………..\sqrt{a_{n+1}} – \sqrt{a_{n}}$
= $\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$ (Multiplying both numerator and denominator by n)
= $\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} – {a_1}}$ $(a_{n+1} – {a_1} =nd)$
= $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$
Question 18: If 8 $\leq$ |2x – 4|*|y – 5| $\leq$ 16, then find the number of order pairs of (x, y) where both x and y are whole numbers.
a) 30
b) 32
c) 33
d) 34
e) 40
18) Answer (E)
Solution:
We are given that, 8 $\leq$ |2x – 4|*|y – 5| $\leq$ 16
4 $\leq$ |x – 2|*|y – 5| $\leq$ 8.
It is given that both x and y are whole numbers. Hence, we can say that x $\geq$ 0. Therefore, x – 2 $\geq$ 2.
Case 1: When |x – 2| = 1, then 4 $\leq$|y – 5| $\leq$ 8
i.e. |x – 2| = 1 and |y – 5| = 4, 5, 6, 7, 8
x = $\pm$1+2 and y = $\pm$4+5, $\pm$5+5, $\pm$6+5, $\pm$7+5, $\pm$8+5
We can see that x can take two values (1, 3) whereas y can take 5 values (1, 9, 0, 10, 11, 12, 13). Hence, a total of 14 values of (x, y) are possible in this case.
Case 2: When |x – 2| = 2, then 2 $\leq$|y – 5| $\leq$ 4
i.e. |x – 2| = 2 and |y – 5| = 2, 3, 4
x = $\pm$2+2 and y = $\pm$2+5, $\pm$3+5, $\pm$4+5
We can see that x can take two values (0, 4) whereas y can take 6 values (3, 7, 2, 8, 1, 9). Hence, a total of 12 values of (x, y) are possible in this case.
Case 3: When |x – 2| = 3, then 4/3 $\leq$|y – 5| $\leq$ 8/3
i.e. |x – 2| = 3 and |y – 5| = 2
x = $\pm$3+2 and y = $\pm$2+5
We can see that x can take only one value whereas y can take two values. Hence, a total of 2 values of (x, y) are possible in this case.
Case 4: When |x – 2| = 4, then 1 $\leq$|y – 5| $\leq$ 2
i.e. |x – 2| = 4 and |y – 5| = 1, 2
x = $\pm$4+2 and y = $\pm$1+5, $\pm$2+5
We can see that x can take only one value whereas y can take four values. Hence, a total of 4 values of (x, y) are possible in this case.
Case 5: When |x – 2| = 5, then 4/5 $\leq$|y – 5| $\leq$ 8/5
i.e. |x – 2| = 5 and |y – 5| = 1
x = $\pm$5+2 and y = $\pm$1+5
We can see that x can take only one value whereas y can take two values. Hence, a total of 2 values of (x, y) are possible in this case.
Case 6: When |x – 2| = 6, then 2/3 $\leq$|y – 5| $\leq$ 4/3
i.e. |x – 2| = 6 and |y – 5| = 1
x = $\pm$6+2 and y = $\pm$1+5
We can see that x can take only one value whereas y can take two values. Hence, a total of 2 values of (x, y) are possible in this case.
Case 7: When |x – 2| = 7, then 4/7 $\leq$|y – 5| $\leq$ 8/7
i.e. |x – 2| = 7 and |y – 5| = 1
x = $\pm$7+2 and y = $\pm$1+5
We can see that x can take only one value whereas y can take two values. Hence, a total of 2 values of (x, y) are possible in this case.
Case 8: When |x – 2| = 8, then 1/2 $\leq$|y – 5| $\leq$ 1
i.e. |x – 2| = 8 and |y – 5| = 1
x = $\pm$8+2 and y = $\pm$1+5
We can see that x can take only one value whereas y can take two values. Hence, a total of 2 values of (x, y) are possible in this case.
Therefore, the total number of possible values of (x, y) = 14+12+2+4+2+2+2+2 = 40. Hence, option E is the correct answer.
Question 19: If a and b are integers such that $2x^2−ax+2>0$ and $x^2−bx+8≥0$ for all real numbers $x$, then the largest possible value of $2a−6b$ is
19) Answer: 36
Solution:
Let f(x) = $2x^2−ax+2$. We can see that f(x) is a quadratic function.
For, f(x) > 0, Discriminant (D) < 0
$\Rightarrow$ $(-a)^2-4*2*2<0$
$\Rightarrow$ (a-4)(a+4)<0
$\Rightarrow$ a $\epsilon$ (-4, 4)
Therefore, integer values that ‘a’ can take = {-3, -2, -1, 0, 1, 2, 3}
Let g(x) = $x^2−bx+8$. We can see that g(x) is also a quadratic function.
For, g(x)≥0, Discriminant (D) $\leq$ 0
$\Rightarrow$ $(-b)^2-4*8*1<0$
$\Rightarrow$ $(b-\sqrt{32})(b+\sqrt{32})<0$
$\Rightarrow$ b $\epsilon$ (-$\sqrt{32}$, $\sqrt{32}$)
Therefore, integer values that ‘b’ can take = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
We have to find out the largest possible value of $2a−6b$. The largest possible value will occur when ‘a’ is maximum and ‘b’ is minimum.
a$_{max}$ = 3, b$_{min}$ = -5
Therefore, the largest possible value of $2a−6b$ = 2*3 – 6*(-5) = 36.
Question 20: If PK: KQ = 3:5, PL: LR = 5:7. Then the ratio of the area of the $\triangle$ KSQ : LSR is
a) 125:98
b) 120:49
c) 120:98
d) 125:49
20) Answer (A)
Solution:
Let us first try to find the ratio of the area of triangles PKS and QKS
$\dfrac{Area (\triangle PKS)}{Area(\triangle QKS)}$ = $\dfrac{\frac{1}{2} * PK * KS * sin PKS} {\frac{1}{2} * QK * KS * (sin 180- PKS) }$
$\dfrac{Area (\triangle PKS)}{Area(\triangle QKS)}$ = $\frac{3}{5}$
Similarly $\dfrac{Area(\triangle PSL)}{Area(\triangle RSL)}$ = $\frac{5}{7}$
$\dfrac{Area(\triangle PKS)}{Area(\triangle QKS)}$ = $\dfrac{Area(\triangle PRK)}{Area(\triangle QRK)}$ [Ratio of the triangles with same alttitudes will be equal to the ratio of bases]
$\dfrac{Area(\triangle PRK)}{Area(\triangle QRK)}$ = $\frac{3}{5}$
$\dfrac{Area(\triangle PQL)}{Area(\triangle RQL)}$ = $\frac{5}{7}$
Let the area of the $\triangle$ PKS = 3a
area of the $\triangle$ QKS = 5a
area of the $\triangle$ PSL = 5b
area of the $\triangle$ RSL = 7b
area of the $\triangle$ QSR = c
$\dfrac{Area(\triangle PRK)}{Area(\triangle QRK)}$ = $\dfrac{area\left(\triangle PKS\right)+area\left(\triangle PSL\right)+area\left(\triangle RSL\right)}{area\left(\triangle QKS\ \right)+area\left(\triangle QSR\right)}$ = $\frac{3a + 5b + 7b}{5a + c}$ = $\frac{3}{5}$
20b = c …….(1)
Similarly, $\dfrac{Area(\triangle PQL)}{Area(\triangle RQL)}$ = $\frac{3a + 5a + 5b}{7b + c}$ = $\frac{5}{7}$
56a = 5c …….(2)
From (1) and (2), 56a = 100b => 14a = 25b
Now we have to find the ratio of the area of the $\triangle$ KSQ: LSR = 5a: 7b
= 125:98
Question 21: The sum of two numbers is 135 and their HCF is 9. How many such pairs of numbers can be formed ?
a) 2
b) 5
c) 4
d) 6
21) Answer (C)
Solution:
Let the two numbers be $9x$ and $9y$, where $x$ and $y$ are co prime numbers.
=> $9x+9y=135$
=> $x+y=\frac{135}{9}=15$
Now, pairs of natural numbers having sum 15 and co prime are = (1,14), (2,13), (4,11), (7,8)
Thus, 4 such pairs can be formed.
=> Ans – (C)
Question 22: A juice seller sells juice mixture which contains mango juice and water in a certain ratio, p:q (where p and q are natural numbers). To increase his profits, the seller adds more water such that the ratio of mango juice and water gets inverted.
To prevent getting caught the seller adds some quantity of mango juice such that the final mixture has the ratio of mango juice and water same as that of the original mixture.
Which of the following can be the possible ratio of water in the original mixture to the water in the final mixture?
a) 4 : 9
b) 9 : 4
c) 3 : 8
d) 1 : 3
22) Answer (A)
Solution:
Let the ratio of mango juice and water in the original mixture be m : w.
Now after adding w1 quantity of water we have,
$\frac{w + w1}{m} = \frac{m}{w}$
Since the value of m and w are natural numbers, the quantity of mango juice and water after adding more water can be m and $\frac{m^2}{w}$ respectively such that the ratio comes out to be w : m.
Now after adding more mango juice the ratio is back to original, thus the quantity of mango juice and water after addition of mango juice can be $\frac{m^3}{w^2}$ and $\frac{m^2}{w}$ respectively such that the ratio comes out to be m : w.
So the ratio of water in the original mixture to the water in the final mixture = $\frac{w^2}{m^2}$
Only option A is in the form of perfect square.
Hence, option A is the right choice.
Although option B is also in the form of a perfect square, the water in the final mixture is greater than the water in the original mixture and hence this is ruled out.
Question 23: While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is
23) Answer: 40
Solution:
We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20
Product of 2 real numbers is 20.
We have to find the minimum possible value of the sum of the squares of the 2 numbers.
Let x=a*b
It has been given that a*b=20
The least possible sum for a given product is obtained when the numbers are as close to each other as possible.
Therefore, when a=b, the value of a and b will be $\sqrt{20}$.
Sum of the squares of the 2 numbers = 20 + 20 = 40.
Therefore, 40 is the correct answer.
Question 24: A shopkeeper marks his books at 25% above the cost price. Due to slump in the market, his cost reduces by 5%. And then,to boost his sale, he offered a discount of 8% due to which sales goes up by 25%. Compute the change in the shopkeepers profit.
a) No change
b) 7% change
c) 2.5% change
d) 8% change
24) Answer (A)
Solution:
Let cost price of each article = Rs. 100 and number of articles sold be $x$
=> Original selling price price = Rs. 125
=> Original profit = Rs. $(125-100)x = Rs.$ $25x$
After offering discount of 8%, => New selling price = $\frac{92}{100}\times125=Rs.$ $115$
Also, new cost price = $\frac{95}{100}\times100=Rs.$ $95$
$\because$ Sale increases by 25%, => Number of articles now sold = $1.25x$
=> New profit = Rs. $(115-95)\times1.25x = Rs.$ $25x$
$\therefore$ There is no change in profit.
=> Ans – (A)
Question 25: A shopkeeper decreases the price of an article by 20% on day 1. On day 2, he increases the price of the article by 20%. He follows this cycle on days 3 and 4, 5 and 6 and so on. Find the day on which the price of the article goes below 60% of the initial price for the first time.
a) 17th day
b) 18th day
c) 19th day
d) 20th day
25) Answer (A)
Solution:
Consider the price to be 100 rs. After decreasing by 20% it became 80 and after increasing by 20% it became 96.
After decreasing and increasing by 20% the price becomes 92.16 Rs.
As we can see, after every 2 days, the price of the article decreases by 4%.
=> $0.96^n$
At n = 8, the value is 0.7211 and the next day the price of the article goes down by 20% => less than 60% of the initial price.
Hence 8*2 + 1 = 17th day
Question 26: A rice trader mixes three types of rice: type A, type B and type C such that the price of type A is more than the price of type B which is more than the price of type C. If x kg of a mixture contains type A, type B and type C in the ratio 1:2:2 and sold at the price of type A, then 66.67% profit is obtained. Now $\ \frac{\ 2x}{5}$ amount of rice of type C is added and the mixture is sold at the price of type B, then only 16.67% profit is obtained. Find the ratio of the per-unit price of type A to that of type B?
a) 7:5
b) 8:5
c) 9:5
d) 10:7
26) Answer (B)
Solution:
Assuming price per kg for type A, type B and type C as a, b and c respectively such that a>b>c.
Assuming x = 5 kg (1+2+2=5, hence to remove fractions x should be a multiple of 5)
Hence, 5 kg of the mixture will contain 1 kg of A, 2 kg of B and 2 kg of C.
The cost price of the mixture per kg = $\ \ \frac{\ a+2b+2c\ }{5}$
Selling price pe kg = a = $\ \frac{\ a+2b+2c}{5}\times\ \left(1+\ \frac{\ 66.67}{100}\right)$ = $\ \frac{\ a+2b+2c}{5}\times\ \ \frac{\ 5}{3}$ = $\ \frac{\ a+2b+2c}{3}$
=> $\ \frac{\ a+2b+2c}{3}$ =a
=> a+2b+2c=3a => a-b=c ……(1)
After adding 2x/5 = 5*2/5 =2 kg of C, we get the cost price of mixture per kg = $\frac{\left(a+2b+4c\right)}{7}$
=> Selling price pe kg = $\ \frac{\ a+2b+4c}{7}\times\ (1+\ \frac{\ 1}{6})$ = b
=> a+2b+4c=6b => 4c= 4b-a …..(2)
Eliminating c from 1 and 2, we get, 4a-4b=4b-a => 5a=8b
=> a/b =8/5
Hence, B is the answer.
Question 27: Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.
27) Answer: 4000
Solution:
Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+Y) = $\dfrac{10}{100}*(X+Y)$ = 0.1(X+Y)
Gopal also borrowed Rs. X from Ankit at 8% per annum. Therefore, he has to return Ankit Rs. 0.08X as the interest amount on borrowed sum.
Hence, the interest retained by gopal = 0.1(X+Y) – 0.08X = 0.02X + 0.1Y … (1)
It is given that the net interest retained by Gopal is the same as that accrued to Ankit.
Therefore, 0.08X = 0.02X + 0.1Y
$\Rightarrow$ X = (5/3)Y … (2)
Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+2Y) = $\dfrac{10}{100}*(X+2Y)$ = 0.1X+0.2Y
In this case the amount of interest retained by Gopal = 0.1X+0.2Y – 0.08X = 0.02X + 0.2Y … (3)
It is given that the interest retained by Gopal increased by Rs. 150 in the second case.
$\Rightarrow$ (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150
$\Rightarrow$ Y = Rs. 1500
By substituting value of Y in equation (2), we can say that X = Rs. 2500
Therefore, (X+Y) = Rs. 4000.
Question 28: Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is
28) Answer: 12
Solution:
Let the distance between A and B be 4x.
Length of BP is thrice the length of AP.
=> AP = x and BP = 3x
Let the speed of car 1 be s and the speed of car 2 be 0.5s.
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.
=> x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60
x/s = 12
Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.
Question 29: 2 friends Arun and Bavin start from 2 cities P and Q towards each other. They meet for the first time when they both of them are at a distance of 60 km from P. After meeting, Arun continues towards Q and Bavin continues towards P. On reaching Q and P, Arun and Bavin return towards the city they started from. They meet for the second time when they are at a distance of 80 km from Q. The ratio of the speeds of Arun and Bavin can be expressed as m:n, where m and n are relatively prime. Enter the value of m+n.
29) Answer: 5
Solution:
Let the distance between the 2 cities be $D$.
Now, when they meet for the first time, Arun would have covered 60 km and Bavin would have covered $D – 60$ Km.
Therefore, the ratio of speeds of Arun and Bavin = $60/(D-60)$————–(1)
When they meet for the second time, Arun would have covered D + 80 Km and Bavin would have covered $(2D – 80)$ Km.
Ratio of speeds of Arun and Bavin = $(D+80)/(2D-80)$ ———————-(2)
Using (1) and (2), we can write,
$60/(D-60) = (D+80)/(2D – 80)$
$ 120D – 4800 = D^2+ 20D – 4800$
$ 120D = D^2 + 20D$
$ D^2 = 100D$
$ D = 100$
Therefore, the ratio of speeds of Arun and Bavin will be 60/40 = 3/2
m + n = 3 + 2 = 5.
Question 30: The owner of an art shop conducts his business in the following manner: every once in a while he raises his prices by X%, then a while later he reduces all the new prices by X%. After one such updown cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?
a) Rs. 2,756.25
b) Rs. 2,256.25
c) Rs. 2,500
d) Rs. 2,000
30) Answer (A)
Solution:
Let the price of the painting be P
One cycle of price increase and decrease reduces the price by $x^2/100 * P = 441$
Let the new price be N => $P – x^2/100 * P = N$
Price after the second cycle = $N – x^2/100 * N$ = 1944.81
=> $(P – x^2/100 * P)(1 – x^2/100) = 1944.81$
=> $(P – 441)(1 – 441/P) = 1944.81$
=> $P – 441 – 441 + 441^2/P = 1944.81$
=> $P^2 – (882 + 1944.81)P + 441^2 = 0$
=> $P^2 – 2826.81P + 441^2 = 0$
From the options, the value 2756.25 satisfies the equation.
So, the price of the article is Rs 2756.25
Question 31: How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
31) Answer: 502
Solution:
It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number.
There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $9C_2$ ways.
3 digit numbers can be formed in $9C_3$ ways.
…………………………………………..
..9 digit number can be formed in 9C9 ways.
We know that $nC_0+nC_1+nC_2+………+nC_n =2^n$
=> $9C_0 + 9C_1+9C_2+….9C_9 = 2^9$
$9C_0 + 9C_1 + …9C_9 = 512$
We have to subtract $9C_0$ and $9C_1$ from both the sides of the equations since we cannot form single digit numbers.
=> $9C_2 + 9C_3+…+9C_9=512-1-9$
$9C_2+9C_3+…+9C_9=502$
Therefore, $502$ is the right answer.
Question 32: During the next year, the probability that a Company A releases a mobile is 0.7. The probability that mobile is a success. given that it is released by the Company is 0.8. The probability that a mobile is a success and released by a Company B is 0.28. A mobile released by either Company A or Company B during the next one year is a success. Find the probability that it is released by Company A.
a) $\frac{1}{3}$
b) $\frac{1}{2}$
c) $\frac{2}{3}$
d) $\frac{3}{4}$
32) Answer (C)
Solution:
Probability that A releases a mobile = 0.7
Probability that mobile is a success given that it is released by A = 0.8
Probability that mobile is released by A and is success = 0.8*0.7 = 0.56
And probability that mobile is a success and released by B = 0.28
Total probability for success = 0.56+0.28 = 0.84
Now Probability that it is success and is released by A = 0.56/0.84 = 2/3
Question 33: John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is
33) Answer: 121000
Solution:
We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years.
John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years.
Let the amount paid as installment every year be Rs.x.
John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid.
=> 1.1x + x = 2,54,100
2.1x = 2,54,100
=> x = Rs. 1,21,000.
Therefore, 121000 is the correct answer.
Question 34: A survey was conducted in a school to find the inclination of the students to read a subject. It was found that 89% students liked Maths, 92% students liked Physics, 85% students liked Chemistry, 74% students liked Biology, 78% students liked Languages and 2% students liked none of the above-mentioned subjects. If the total number of students in the school is 400, then the minimum number of students who liked all the five subjects is
(Enter ‘-1’ if your answer is Cannot be determined )
34) Answer: 104
Solution:
The percentage of students in school=100%
Percentage of students who like none of the above mentioned subjects=2%
Consider the number of students who like exactly 1 subject = I
The number of students who like exactly 2 subjects = II
The number of students who like exactly 3 subjects = III
The number of students who like exactly 4 subjects = IV
The number of students who like exactly 5 subjects = V
I+II+III+IV+V=98
I+2II+3III+4IV+5V=89+92+85+74+78=418
Since we have to find the minimum number of students who like all the five subjects, we have to maximize the number of students who like four subjects.
So the rest of the variables can be zero.
4IV+5V=418
IV+V=98
V=26% and IV = 72%
Here we got the minimum percentage of students who like all the five subjects.
In this case, 72% of the students like exactly 4 subjects each and 26% students like all the five subjects. The venn diagram for such a case is given below
To find the number of students, we multiply it with the total number of students and hence it equals = 400*26/100 = 104
Question 35: For any real x, the functions f(x) and g(x) are defined as f(x) = $\ \dfrac{\ 20x}{1+x^2}$ and g(x) = $\ \dfrac{\ 2}{4x-2x^2-1}$, then how many positive integral values are common between the total values f(x) and g(x) can take?
a) 7
b) 8
c) 9
d) 10
35) Answer (C)
Solution:
We have, $\ \dfrac{\ 20x}{1+x^2}$ = $\ \dfrac{\ 20}{\frac{1}{x}+x}$
The value of $x+\frac{1}{x}$ can range from (-$\infty\ $,-2] U [2,$\infty\ $) (Using AM-GM)
Then the value of $\ \dfrac{\ 20x}{1+x^2}$ will be $\left[\frac{20}{-2},\frac{20}{2}\right]$ = [-10,10] ( f(x) can take value of 0 at x=0)
Also, g(x) = $\ \dfrac{\ 2}{4x-2x^2-1}$
To find minimum or maximum value, the quadratic function can be maximized or minimized.
Hence, g(x) = $\ \dfrac{\ 2}{4x-2x^2-1}$ = $\ \dfrac{\ 1}{-\frac{1}{2}-x^2+2x}$ = $\ \dfrac{\ 1}{-\frac{1}{2}-x^2+2x-1+1}$=$\ \dfrac{\ 1}{-\frac{1}{2}-\left(x-1\right)^2+1}$ = $\ \dfrac{\ 1}{\frac{1}{2}-\left(x-1\right)^2}$
Hence, the minimum positive value of g(x) is obtained when the square value is minimum which can be zero.
Hence the minimum value = $\ \dfrac{\ 1}{\frac{1}{2}-0}=2$ and the maximum value can be infinity when the denominator approaches 0.
So the positive integers, the value of which both f(x) and g(x) can take will be 2,3,4,5,6,7,8,9,10. Hence a total of 9 positive integral values are common.
Alternate Solution:
Assume y = $\ \dfrac{\ 20x}{1+x^2}$
=> $x^2-\frac{20x}{y}+1=0$
For x to be real, the discriminant of the the equation should be greater than 0.
Hence, $\left(\frac{20}{y}\right)^2-4\ge\ 0$
=> $\frac{\left(100-y^2\right)}{y^2}\ge\ 0$
=> $y^2-100\le\ 0$ ($y^2$ would not affect the sign)
=> y $\in\ \left[-10,10\right]$ [y can take the value 0 at x=0]
Similarly for g(x), assume y = $\ \dfrac{\ 2}{4x-2x^2-1}$
=> $-2x^2+4x-1=\frac{2}{y}$
=> $2x^2-4x+1+\frac{2}{y}=0$
For x to be real, the discriminant of the the equation should be greater than 0.
Hence, $16-4\times\ 2\times\ \left(1+\frac{2}{y}\right)\ge\ 0$
=> $1-\frac{2}{y}\ge\ 0$
=> $\frac{y-2}{y}\ge\ 0$
=> y $\in\ \left(-\infty,\ 0\right)U\left[2,\ \infty\right)$
Hence, the common positive integral between two functions are 2,3,4,5,6,7,8,9,10. => 9 positive integral values.
Question 36: Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is
a) 53
b) 51
c) 48
d) 49
36) Answer (B)
Solution:
Assume the average of 21 students other than Ramesh = a
Sum of the scores of 21 students other than Ramesh = 21a
Hence the average of 22 students = a+1
Sum of the scores of all 22 students = 22(a+1)
The score of Ramesh = Sum of scores of all 22 students – Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22 = 82.5 (Given)
=> a = 60.5
Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353
Now the sum of the scores of students other than Gautam = 21*62 = 1302
Hence the score of Gautam = 1353-1302=51
Question 37: The minimum possible value of the sum of the squares of the roots of the equation $x^2+(a+3)x-(a+5)=0 $ is
a) 1
b) 2
c) 3
d) 4
37) Answer (C)
Solution:
Let the roots of the equation $x^2+(a+3)x-(a+5)=0 $ be equal to $p,q$
Hence, $p+q = -(a+3)$ and $p \times q = -(a+5)$
Therefore, $p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$
As $(a+4)^2$ is always non negative, the least value of the sum of squares is 3
Question 38: Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is
38) Answer: 5
Solution:
Time taken to cover first 50 km at 100 km/hr = $\frac{1}{2}$ hr.
Time taken to cover second 50 km at 50 km/hr = 1 hr.
Time taken to cover last 50 km at 25 km/hr = 2 hr.
When car 2 starts, car 1 has already covered 20 km.
So, time taken by car 1 to reach B after car 1 starts = total time – time required to travel first 20 km
= 3 hr 30 min – 12 min = 3 hr 18 min
Distance travelled by car 2 = (50 + 50 + 45) = 145 km
Distance from B = (150 – 145) km = 5 km
Hence, 5 is the correct answer.
Question 39: If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be
a) $(1003)^{15} + 6$
b) $(997)^{15} – 3$
c) $(997)2^{14} + 3$
d) $(1003)2^{15} – 3$
39) Answer (D)
Solution:
The population of town at the beginning of 1st year = p
The population of town at the beginning of 2nd year = 3+2p
The population of town at the beginning of 3rd year = 2(3+2p)+3 = 2*2p+2*3+3 =4p+3(1+2)
The population of town at the beginning of 4th year = 2(2*2p+2*3+3)+3 = 8p+3(1+2+4)
Similarly population at the beginning of the nth year = $2^{n-1}$p+3($2^{n-1}-1$) = $2^{n-1}\left(p+3\right)$-3
The population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be $(2^{2034-2019})\left(1000+3\right)$-3 = $2^{15}\left(1003\right)$-3
Question 40: In $\triangle ABC$, points D, E, and F divide the sides AC, AB and BC respectively such that $AE:EB=BF:FC=CD:DA=3:1$. Similarly, points G, H, and I divide the sides DE, EF and FD such that $DG:GE=EH:HF=FI:ID=1:2$. Find the ratio of the area of $\triangle ABC$ and $\triangle GHI$.
a) 48:7
b) 256:49
c) 9:1
d) 43:12
40) Answer (A)
Solution:
Let us consider $\triangle AED$ & $\triangle ABC$
$\frac{area(\triangle AED)}{area(\triangle ABC)}=\frac{\frac{1}{2}\times AE\times AD\times sinA}{\frac{1}{2}\times AB\times AC\times sinA}=\frac{\frac{1}{2}\times 3x \times z \times sinA}{\frac{1}{2}\times 4x \times 4z\times sinA}=\frac{3}{16}$
Similarly, $\frac{area(\triangle BEF)}{area(\triangle ABC)}=\frac{3}{16}$
and $\frac{area(\triangle CFD)}{area(\triangle ABC)}=\frac{3}{16}$
So, if $area(\triangle ABC)=16k$
then, $area(\triangle DEF)=16k-9k=7k$
Now, let us consider $\triangle DGI$ and $\triangle DEF$
$\frac{area(\triangle DGI)}{area(\triangle DEF)}=\frac{\frac{1}{2}\times DG\times DI\times sinD}{\frac{1}{2}\times DE\times DF\times sinAD}=\frac{\frac{1}{2}\times p \times 2r \times sinD}{\frac{1}{2}\times 3p \times 3r\times sinD}=\frac{2}{9}$
Similarly, $\frac{area(\triangle GEH)}{area(\triangle DEF)}=\frac{2}{9}$
and $\frac{area(\triangle FHI)}{area(\triangle DEF)}=\frac{2}{9}$
So, $area(\triangle GHI)=(1-\frac{6}{9})\times 7k=\frac{1}{3} \times 7k=\frac{7k}{3}$
$\therefore \frac{area(\triangle ABC)}{area(\triangle GHI)}=\frac{16k}{\frac{7k}{3}}=\frac{48}{7}$
Hence, option A is the correct answer.
Question 41: If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be
a) 23
b) 26
c) 96
d) 93
41) Answer (D)
Solution:
It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values.
The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too.
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.
The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible.
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95.
The number of students who like only burger should lie between 29 and 95 (both the values are included).
93 is the only value among the given options that satisfies this condition and hence, option D is the right answer.
Question 42: The numbers 1, 2, …, 9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.
If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is
42) Answer: 3
Solution:
According to the question each column, each row, and each of the two diagonals of the 3X3 matrix add up to the same value. This value must be 15.
Let us consider the matrix as shown below:
Now we’ll try substituting values from 1 to 9 in the exact middle grid shown as ‘x’.
If x = 1 or 3, then the value in the left bottom grid will be more than 9 which is not possible.
x cannot be equal to 2.
If x = 4, value in the left bottom grid will be 9. But then addition of first column will come out to be more than 15. Hence, not possible.
If x=5, we get the grid as shown below:
Hence, for x = 5 all conditions are satisfied. We see that the bottom middle entry is 3.
Hence, 3 is the correct answer.
Question 43: A shopkeeper sells tomatoes. He buys 100 kg of tomatoes and sells them in packages of 1 kg and 3 kg. He offers 2 packets of 1 kg of tomato free for customers who buy the 3 kg packet. He marks up the price of the tomatoes by 30% and ends up realizing a profit percentage of 17%. What percentage of the total quantity of tomatoes sold are sold in 3 kg packets?
a) 25%
b) 30%
c) 15%
d) 35%
43) Answer (C)
Solution:
Let the cost price of 1 kg of tomato be Rs.1.
Selling price of 1 kg tomato packet = Rs.1.3
Selling price of 3kg of tomato packet = Rs.3.9.
But, by paying Rs. 3.9, the customer gets 5 kg of tomatoes.
Therefore, the selling price of tomatoes that are not sold as 1 kg packets = 3.9/5
Let the number of 1 kg packets that are sold for a price (not counting the ones given away with the 3 kg packets) be x.
x kg of tomatoes are sold as 1 kg packets and the remaining are sold as 5 kg combos (3 kg + 2*1kg packets).
It has been given that the net profit percentage is 17%.
Therefore, the net revenue = Rs. 117
1.3x + 3.9(100-x)/5 = 117
6.5x + 390 – 3.9x = 585
2.6x = 195
x = 75
Therefore, 75 kg of tomatoes are sold as 1 kg packets.
Remaining 25 kg must be sold as 5 kg combos.
Therefore, 25/5 = 5 combos are sold in total.
5 combos will contain 5 3 kg packets.
Therefore, quantity of tomato sold as 3 kg packets = 5*3 = 15 kg or 15%.
Therefore, option C is the right answer.
Question 44: Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is
44) Answer: 9
Solution:
Area of ABD : Area of BDC = 1:1
Therefore, area of ABD = 54
Area of ADE : Area of EDB = 1:1
Therefore, area of ADE = 27
O is the centroid and it divides the medians in the ratio of 2:1
Area of BEO : Area of EOD = 2:1
Area of EOD = 9
Question 45: In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is
a) 2.5
b) 1.5
c) 3.5
d) 0.5
45) Answer (D)
Solution:
In figure AE*BE=CE*DE (The intersecting chords theorem)
=> 7*15= x(20.5-x) (Assuming AE=x)
=> 210=x(41-2x)
=> $2x^2$-41x+210=0
=> x=10 or x=10.5 => AE=10 or AE=10.5 Hence BE = 20.5-10=10.5 or BE = 20.5-10.5=10
Required difference= 10.5-10=0.5
Question 46: A person earned an interest of ‘S’ when he invested a principal at simple interest for 3 years. He earned an interest of ‘C’ when he invested the same principal at compound interest for 2 years. The rate of interest was same in both the cases. If the ratio S:C = 15:11, then which of the following can be the rate of interest?
a) 5%
b) 10%
c) 15%
d) 20%
e) 25%
46) Answer (D)
Solution:
Let the rate of interest be ‘100r %’, Principal be ‘P’.
S = P*3*r = 3Pr
C = $P(1+r)^2-P=P(r^2+2r)$
S:C = 15:11
=> $\frac{3Pr}{P(r^2+2r)}=\frac{15}{11}$
=> $15r^2+30r=33r$
=> r = 0.2
Rate of interest is 100 X .2 = 20%.
Question 47: The smallest integer $n$ such that $n^3-11n^2+32n-28>0$ is
47) Answer: 8
Solution:
We can see that at n = 2, $n^3-11n^2+32n-28=0$ i.e. (n-2) is a factor of $n^3-11n^2+32n-28$
$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$
We can further factorize n^2-9n+14 as (n-2)(n-7).
$n^3-11n^2+32n-28=(n-2)^2(n-7)$
$\Rightarrow$ $n^3-11n^2+32n-28>0$
$\Rightarrow$ $(n-2)^2(n-7)>0$
Therefore, we can say that n-7>0
Hence, n$_{min}$ = 8
Question 48: Let $K$ be a natural number such that the divisors of $10^K$ are $a_1,a_2,….,a_p$. If it is known that the value of $\log _{10} a_1 + \log_{10} a_2 …. + \log_{10} a_p = 605$. What is the value of $K$
48) Answer: 10
Solution:
$\log _{10} a_1 + \log_{10} a_2 …. + \log_{10} a_p = 605$
$\log _{10} {(a_1 * a_2 …. * a_p)} = 605$
Hence, $a_1 * a_2 *…. a_p = 10^{605}$
$10^K = 2^K \times 5^K$
Hence, the number of divisors of $10^K$ is $(K+1) \times (K+1)$
So, the value of $p=(K+1)^2$
If we assume that $a_1,a_2,….a_p$ are arranged in ascending order, the value of $a_1*a_p = 10^K$. Similarly, the value of $a_2*a_{p-1}=10^K$ and so on.
If $p$ is odd, say $(2b+1)$, the value of $a_1*a_2*…a_p = (a_1*a_p) * (a_2*a_{p-1}) * …. * a_{b} = 10^K *10^K … $(b times)$*10^{K/2}=10^{Kb+K/2} = 10^{K(2b+1)/2} = 10^{Kp/2}$
If p is even, say (2b), the value of $a_1*a_2*…a_p = (a_1*a_p) * (a_2*a_{p-1}) * …. = 10^K *10^K … $(b times) = $10^{Kb/2} = 10^{Kp/2}$
In both the cases, the value of $a_1*a_2*…a_p = 10^{Kp/2} = 10 ^ {K*(K+1)^2/2}$
Therefore, the value of $K*(K+1)^2/2 = 605$ or
$K*(K+1)^2 = 1210$
As K is a naural number, the value of $K = 10$
Question 49: The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is
49) Answer: 92
Solution:
Let sum of marks of students be x
Now therefore x = 25*50 =1250
Now to maximize the marks of the toppers
We will minimize the marks of 20 students
so their scores will be (30,31,32…..49 )
let score of toppers be y
so we get 5y +$\frac{20}{2}\left(79\right)$=1250
we get 5y +790=1250
5y=460
y=92
So scores of toppers = 92
Question 50: One spherical ball is placed over a hollow cylindrical box. The radius of the sphere is 14 cm, while the cylinder is of radius 7 cm. If the height of the cylinder is 10 cm, what is the total height of the figure?
a) 24+7$\sqrt{\ 3}$ cm
b) 24 cm
c) 14+7$\sqrt{\ 3}$ cm
d) 24+14$\sqrt{\ 3}$ cm
50) Answer (A)
Solution:
The figure can be represented as :
The triangle formed is an equilateral triangle with the side of length 14cm
The height of an equilateral triangle= $\frac{\sqrt{\ 3}}{2}\cdot a$
=$\frac{\sqrt{\ 3}}{2}\cdot 14$
= $7\sqrt{\ 3}$
Height = 10+ $7\sqrt{\ 3}$+ 14 cm
= 24+$7\sqrt{\ 3}$
Option A is the correct answer.
Question 51: In an examination containing 100 questions, a student is awarded 3 marks for a correct answer, -2 for a wrong answer and -1 for an un-attempted question. How many different scores are possible for a student who takes this examination?
a) 495
b) 497
c) 499
d) 501
51) Answer (A)
Solution:
Let the number of correct answers, number of wrong answers and number of un-attempted questions be a, b and c respectively.
The student gets 3 marks for a correct, -2 for a wrong answer and -1 for an un-attempted question => Total marks are a function of (3a – 2b – c).
We know that a + b + c = 100 as the total number of questions is 100.
So, the range of (3a – 2b – c) will be between -200 and 300. But not all values in this range are possible.
First consider the negative score values. The least score possible is -200, when all answers are wrong. -199 is attained by 99 wrongs and 1 unattempted question. -198 is attained by 98 wrongs and 2 unattempted questions. This linear relationship goes on until the score of -100, which is attained by 100 unattempted questions. Now, for a score of -99, one needs at least one correct answer. So assuming 1 question is correct, which gives +3, the student has to get -102 in remaining questions to get an overall score of -99. -102 is possible by 96 unattempted and 3 wrongs. On increasing the number of correct answers, one can score all the values from -99 to -1, which is attained at 33 correct answers, 33 wrong answers and 34 unattempted questions. Clearly, we can see that all the negative scores until -200 are possible.
Now, consider the positive score values that are attained by (3a – 2b – c). 300 is possible if a is 100. But 299, 298 and 297 are only possible if a is 100 and b and c are positive integers. But, this is not possible as the sum of a, b and c cannot exceed 100.
Similarly, the values 294, 293 and 289 cannot be formed by the expression (3a – 2b – c). Therefore, in total 6 scores are not possible, which gives us 501 – 6 = 495 different scores.
Question 52: In an examination, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali’s score exceeded Rama’s score by
a) 26
b) 32
c) 35
d) 24
52) Answer (B)
Solution:
Let the scores of Rama, Anjali and Mohan be r, a, m.
It is given that Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali
r=$\ \frac{\ m+a}{12}$ ———-(1)
The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6
a+6:m+6:r+6 = 11:10:3
Let a+6 = 11x => a= 11x-6
m+6=10x => m=10x-6
r+ 6 =3x => r = 3x-6
Substituting these values in equation (1), we get
3x-6=$\ \frac{\ 21x-12}{12}$
12(3x-6) = 21x-12
x=4
Anjali’s score exceeds Rama’s score by (a-r)=8x=32
Question 53: How many factors of $2^4 \times 3^5 \times 10^4$ are perfect squares which are greater than 1?
53) Answer: 44
Solution:
$2^4 \times 3^5 \times 10^4$
=$2^4 \times 3^5 \times 2^4*5^4$
=$2^8 \times 3^5 \times 5^4$
For the factor to be a perfect square, the factor should be even power of the number.
In $2^8$, the factors which are perfect squares are $2^0, 2^2, 2^4, 2^6, 2^8$ = 5
Similarly, in $3^5$, the factors which are perfect squares are $3^0, 3^2, 3^4$ = 3
In $5^4$, the factors which are perfect squares are $5^0, 5^2, 5^4$ = 3
Number of perfect squares greater than 1 = 5*3*3-1
=44
Question 54: If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures that she makes a profit of 20%?
a) Give a discount of 5% on retail price.
b) Give a discount of 2% on retail price.
c) Increase the retail price by 2%.
d) Sell at retail price.
54) Answer (D)
Solution:
Let the retail price be M and cost price be C.
Given,
0.85 M = 1.02 C
M = 1.2 C
If he wants 20% profit he has to sell at 1.2C, which is nothing but the retail price.
Question 55: A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is
a) $1026(1 + \pi)$
b) $8464\pi$
c) $928\pi$
d) $1044(4 + \pi)$
55) Answer (A)
Solution:
It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)
=27
The number of iron cylinders = $\ \frac{\ 405}{27}$ = 15
The number of aluminium cylinders = $\ \frac{\ 783}{27}$ = 29
The number of copper cylinders = $\ \frac{351\ }{27}$ = 13
15*$\pi\ r^2h=$ 405
15*$\pi\ 9*h=$ 405
$\pi\ h=3$
Now we have to calculate the total surface area of all the cylinders
Total number of cylinders = 15+29+13 = 57
Total surface area of the cylinder = 57*($2\pi\ rh+2\pi\ r^2$)
=57(2*3*3 + 2*9*$\pi\ $)
=$1026(1 + \pi)$
Question 56: With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y+1) or to (x+1, y), is
56) Answer: 3920
Solution:
The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)
To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction
Hence the number of paths from (1,1) to (4,6) = $^{(3+5)}C_3$ = 56
To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction
Hence the number of paths from (4,6) to (8,10) = $^{(4+4)}C_4$ = 70
The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920
Question 57: $\frac{log (97-56\sqrt{3})}{log \sqrt{7+4\sqrt{3}}}$ equals which of the following?
a) None of the others
b) -2
c) -4
d) -3
e) -5
57) Answer (C)
Solution:
Simplify the expression a bit to remove the root sign in the denominator
$\dfrac{\log{97-56\sqrt{3}}}{\dfrac{1}{2}\times (\log{7+4\sqrt{3})}}$
$ \Rightarrow 2 \times \dfrac{\log{97-56\sqrt{3}}}{\log{7+4\sqrt{3}}} $
To move further, let us see the root of the numerator.
Assume the root of the numberator to be $\sqrt{a}-\sqrt{b}$.
When we square it, we get $a + b – (2 \times \sqrt{a} \sqrt{b}) = a+b-2\sqrt{ab} $
comparing the value of terms under root with the terms in the numerator, we get
$\sqrt{ab} = 28\sqrt{3} $ and $a+b=97$
From solving this, we get to know that $a=7$ and $b=4\sqrt{3}$
Thus the expression can be written as $ 2 \times2 \times \dfrac{\log{7-4\sqrt{3}}}{\log{7+4\sqrt{3}}} $
$ \Rightarrow 4 \times \dfrac{\log{7-4\sqrt{3}}}{\log{7+4\sqrt{3}}} $
Now, let us look at the reciprocal of the term in log in the denominator.
$ \frac{1}{7+4\sqrt{\ 3}} = \frac{1}{7+4\sqrt{\ 3}} \times \dfrac{7-4\sqrt{3}}{7-4\sqrt{3}} $
$ \Rightarrow \dfrac{7-4\sqrt{3}}{7^{2}-(4\sqrt{3})^{2}} $
$ \Rightarrow \dfrac{7-4\sqrt{3}}{49-48} = 7-4\sqrt{3}$
$\frac{\log\left(7-4\sqrt{\ 3}\right)}{\log\left(7+4\sqrt{\ 3}\right)}=\frac{\log\left(7-4\sqrt{\ 3}\right)}{\log\left(\frac{1}{7-4\sqrt{\ 3}}\right)}=\frac{\log\left(7-4\sqrt{\ 3}\right)}{-\log\left(7-4\sqrt{\ 3}\right)}=-1$
Thus the value of the expression can be further simplified as
$ \Rightarrow 4 \times (-1) = -4 $
Hence the correct answer is option C
Question 58: A tank of capacity 1200 litres is being filled by two taps A and B. A can fill the tank in 30 hours and B can fill the tank in 40 hours. Due to extended usage, the tank develops a leak at the bottom. However, due to salt deposition at the site of the leak, the size of the hole keeps decreasing uniformly over time. At the start, water leaks from the hole at the rate of half a liter per minute. After 4 hours, this reduces to 100ml per minute and becomes constant thereafter. If both taps are opened, how long will it take to fill the tank?
a) 22.44 hours
b) 15.50 hours
c) 18.24 hours
d) 19.50 hours
58) Answer (D)
Solution:
Lets convert leak timings to litres per hour. Initial rate is 0.5 litres per min = 0.5* 60 = 30 lit / hr
Final rate = 0.1*60 = 6 lit / hr
The rate of leaking can be represented by a graph as follows:
The volume of water leaked is given by the area under the graph. In the first four hours, the area is as indicated by the light blue trapezium with sides 30 and 6 and height 4.
Hence, water leaked = 1/2* (30+6) * 4 = 72 litres
Tap A fills the tank at 1200/30 = 40 lit / hr
Tap B fills the tank at 1200/40 = 30 lit / hr
In the first four hours, the tank is filled up to level = 4*(40+30) – 72 = 4*70 – 72 = 208 litres.
After that, rate of filling the tank = 40+ 30 – 6 = 64 lit / hr
Time taken = (1200 – 208) / 64 = 992/64 = 15.5 hours
Thus, total time taken = 15.5 + 4 = 19.5 hours.
Question 59: When opening his fruit shop for the day a shopkeeper found that his stock of apples could be perfectly arranged in a complete triangular array: that is, every row with one apple more than the row immediately above, going all the way up ending with a single apple at the top.
During any sales transaction, apples are always picked from the uppermost row, and going below only when that row is exhausted.
When one customer walked in the middle of the day she found an incomplete array in display having 126 apples totally. How many rows of apples (complete and incomplete) were seen by this customer? (Assume that the initial stock did not exceed 150 apples.)
a) 15
b) 14
c) 13
d) 12
e) 11
59) Answer (D)
Solution:
The stack of apples can be imagined as shown below. For every row, there is an increase of 1 apple from the previous row.
This means that the sum total of the apples present in the stack will be sum of AP given as : 1,2,3,4…. ie the sum of numbers from 1.
This is given by the formula S = $\frac{n \times (n+1)}{2}$
Since the initial number of apples was not more than 150, we find the maximum number of rows that was possible.
Let S be the total initial sum of apples in the cart.
We know that $S\leq 150 $
$ \Rightarrow \frac{n \times (n+1)}{2} \leq 150 $
$ \Rightarrow n \times (n+1) \leq 300 $
$ \Rightarrow n^2 + n \leq 300 $
Since the biggest number with square less than 300 is 17, we try to see if 17 works in the expression by hit and trial method.
So, 17 x 18 = 306 > 300, therefore the maximum number of rows possible is not 17.
Since 306 is JUST over 300, let us check at 16.
So, 16 x 17 = 272 < 300, therefore the maximum number of rows possible is 16.
Let us tabulate the total number of apples present in the cart based on the number of rows :
Since the total number of apples from row 15 is less than 126 (which was found by the customer), we can safely say that the total number of rows of apples with the shopkeeper that day was 16.
Now, if ‘n’ rows of apples are sold, the total available apples can be calculated as : $\frac{16 \times 17}{2} – \frac{n \times (n+1)}{2}$
Let us tabulate the result obtained in this case :
From the table we can see that when the customer saw 126 apples, the apples in the top 4 rows had been sold.
Therefore, the number of rows = 16 – 4 = 12 rows.
Question 60: A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?
a) 22
b) 23
c) 15
d) 20
60) Answer (C)
Solution:
It is given that starting from 10:01 am, every minute a motorcycle leaves A and moves towards B.
Forty-five such motorcycles reach B by 11 am.
It means that the forty-fifth motorcycle starts at 10:45 AM at A and reaches B by 11:00 AM i.e 15 minutes.
Since the speed of all the motorcycles is the same, all the motorcycles will take the same duration i.e 15 minutes.
If the cyclist doubles the speed, then he will reach B by 10:30 AM. (Since if the speed is doubled, time is reduced by half)
Since each motorcycle takes 15 minutes to reach B, 15 motorcycles would have reached B by the time the cyclist reaches B
Question 61: The product of the distinct roots of $\mid x^2 – x – 6 \mid = x + 2$ is
a) −16
b) -4
c) -24
d) -8
61) Answer (A)
Solution:
We have, $\mid x^2 – x – 6 \mid = x + 2$
=> |(x-3)(x+2)|=x+2
For x<-2, (3-x)(-x-2)=x+2
=> x-3=1 =>x=4 (Rejected as x<-2)
For -2$\le\ $x<3, (3-x)(x+2)=x+2 =>x=2,-2
For x$\ge\ $3, (x-3)(x+2)=x+2 =>x=4
Hence the product =4*-2*2=-16
Question 62: A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is $9\pi$ cubic centimeters. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is
62) Answer: 6
Solution:
The volume of a cylinder is $\pi * r^2 * 3 = 9 \pi$
r = $\sqrt{3}$ cm.
Radius of the ball is 2 cm. Hence, the ball will lie on top of the cylinder.
Lets draw the figure.
Based on the Pythagoras theorem, the other leg will be 1 cm.
Thus, the height will be 3 + 1 + 2 = 6 cm
Question 63: ABCD is a quadrilateral such that its diagonals are perpendicular to each other. If AB=75 and CD = 40. If AD:BC= 3:4, then the sum of the lengths of AD and BC is
63) Answer: 119
Solution:
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Assume the diagonals intersect at O. Now OA$^{2}$+OB$^{2}$+OC$^{2}$+OD$^{2}$ = OA$^{2}$ +OD$^{2}$+OB$^{2}$+OC$^{2}$
Hence AB$^{2}$+CD$^{2}$=AD$^{2}$+BC$^{2}$….(1)
75$^{2}$+40$^{2}$=(3x)$^{2}$+(4x)$^{2}$ (Consider AD=3x and BC=4x)
=> 7225 = 25x$^{2}$
=> x$^{2}$=289
=> x=17
AD+BC=3x+4x=7x=7*17=119
Question 64: When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?
a) 20
b) 22
c) 16
d) 18
64) Answer (A)
Solution:
Let us assume that A can complete ‘a’ units of work in a day and B can complete ‘b’ units of work in a day.
A works alone till half the work is completed.
A and B work together for 4 days and B works alone to complete the last 5% of the work.
=> A and B in 4 days can complete 45% of the work.
Let us assume the total amount of work to be done to be 100 units.
4a + 4b = 45 ———(1)
B needs 25% more time than A to finish a job.
=> 1.25*b = a ———-(2)
Substituting (2) in (1), we get,
5b+4b = 45
9b = 45
b = 5 units/day
B alone can finish the job in 100/5 = 20 days.
Therefore, option A is the right answer.
Question 65: Consider a function f satisfying f (x + y) = f (x) f (y) where x,y are positive integers, and f(1) = 2. If f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1) then a is equal to
65) Answer: 3
Solution:
f (x + y) = f (x) f (y)
Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4
f(3)=f(2+1)=f(2)*f(1)=4*2=8
f(4)=f(3+1)=f(3)*f(1)=8*2=16
…….=> f(x)=$2^x$
Now, f(a + 1) +f (a + 2) + … + f(a + n) = 16 (2$^n$ – 1)
On putting n=1 in the equation we get, f(a+1)=16 => f(a)*f(1)=16 (It is given that f (x + y) = f (x) f (y))
=> $2^a$*2=16
=> a=3
Question 66: The work done by a man, woman and child are in the ratio 3: 2:1. If daily wages of 20 men, 30 women and 36 children amount to 78/-, what will be the wages of 15 men, 21 women and 30 children for 18 weeks ?
a) 7371/-
b) 8645/-
c) 7532/-
d) 9000/-
66) Answer (A)
Solution:
Let the efficiency of a man, woman and child be $3x,2x,x$ espectively.
Thus, 1 day wage = $(3x\times20)+(2x\times30)+(x\times36)=78$
=> $156x=78$
=> $x=\frac{1}{2}$
$\therefore$ Wages of 15 men, 21 women and 30 children for 18 weeks = $[(15\times3x)+(21\times2x)+(30\times x)]\times18\times7$
= $117\times\frac{1}{2}\times126$
= $117\times63=Rs.$ $7,371$
=> Ans – (A)
Question 67: A, B, C, D and E are five employees working in a company. In two successive years, each of them got hikes in his salary as follows:
A : p% and (p+1)%,
B : (p+2)% and (p-1)%,
C : (p+3)% and (p-2)%,
D : (p+4)% and (p-3)%,
E : (p+5)% and (p-4)%.
If all of them have the same salary at the end of two years, who got the least hike in his salary?
a) E
b) D
c) C
d) A
e) B
67) Answer (A)
Solution:
Let the initial salary of A,B,C,D,E be $a,b,c,d,e$ respectively and let the final salary of everyone be $x$
Now, $a*(1+ \frac{p}{100})*(1+ \frac{p+1}{100}) = x$
$\Rightarrow a = \frac{x}{(1+ \frac{p}{100})*(1+ \frac{p+1}{100})}$
$\Rightarrow a = \frac{x*100*100}{(100+p)*(100+p+1)}$
$\Rightarrow a = \frac{x*100*100}{(p+100)*(p+101)}$
Similarly, $b= \frac{x}{(1+ \frac{p+2}{100})*(1+ \frac{p-1}{100})}$
$\Rightarrow b = \frac{x*100*100}{(p+102)*(p+99)}$
Similarly, $c= \frac{x}{(1+ \frac{p+3}{100})*(1+ \frac{p-2}{100})}$
$\Rightarrow c = \frac{x*100*100}{(p+103)*(p+98)}$
Similarly, $d= \frac{x}{(1+ \frac{p+4}{100})*(1+ \frac{p-3}{100})}$
$\Rightarrow d = \frac{x*100*100}{(p+104)*(p+97)}$
Similarly, $e= \frac{x}{(1+ \frac{p+5}{100})*(1+ \frac{p-4}{100})}$
$\Rightarrow e = \frac{x*100*100}{(p+105)*(p+96)}$
The numerators of the fractions are same, therefore the one with the smallest value of denominator will have the greatest value. If we compare the denominators, we can find out the fraction with the highest value. The person with the highest initial salary got the least raise, as we know that the final salary of all the candidates is same.
Thus, denominator of a, $a_{den}= (p+100)*(p+101)= p^{2}+201p+100*101$
Similarly, $b_{den}= (p+102)*(p+99)= p^{2}+201p+102*99$
$c_{den}= (p+103)*(p+98)= p^{2}+201p+103*98$
$d_{den}= (p+104)*(p+97)= p^{2}+201p+104*97$
$e_{den}= (p+105)*(p+96)= p^{2}+201p+105*96$
We see that we need to compare only the last terms of the denominators as the other terms are same.
Thus, last term of a, $a_{lt}= 100*101= 100.5^{2}-0.5^{2}$
last term of b, $b_{lt}= 102*99= 100.5^{2}-1.5^{2}$
last term of c, $c_{lt}= 103*98= 100.5^{2}-2.5^{2}$
last term of d, $d_{lt}= 104*97= 100.5^{2}-3.5^{2}$
last term of e, $d_{lt}= 105*96= 100.5^{2}-4.5^{2}$
Thus, we can see that since denominator of $e$ is the smallest, therefore E has the highest initial salary.
Question 68: A circle with its centre in 1st quadrant touches the x-axis only at x=5, what should be the radius(units) of the circle such that it touches 4x-3y=0
a) 5
b) 1.25
c) 2.5
d) 1.5
68) Answer (C)
Solution:
The given information can be represented as the figure below:
Now, the radius = k (from the figure)
Also, the distance of the centre (3,k) from 4x-3y=0 will be equal to the radius = k.
Hence, k = $\frac{\left|\left(4\times\ 5-3k\right)\right|}{\sqrt{4^2+3^2}}$ (Using the formula for perpendicular distance of a line from a point.
=> $\pm 5k = 20-3k $
=> $ 3k \pm 5k = 20 $
=> k = 2.5 or -10
After rejecting the negative value, k = 2.5 units.
Question 69: David buys a certain number of pencils. He decides to sell the pencils in packets consisting of ‘n’ pencils, earning a profit of 20%. Had David packed ‘n-1’ pencils per packet and sold the packets at the same price, the profit he would have earned is Rs P. If P is 20% more than the amount of profit(Rs) made by selling n pencils per packet, what is the least number of pencils that David could have bought?
a) 930
b) 900
c) 960
d) 620
69) Answer (A)
Solution:
Let us assume that $T$ is the total cost incurred by David in buying the pencils.
Let the total number of pencils be $X$.
Now, the number of packets will be $\frac{X}{n}$.
Let the cost of a packet be $c$.
Total revenue = Number of packets*cost of each packet.
=> Total revenue = $\frac{X}{n}*c$
We know that David earns a profit of 20% by selling ‘n’ pencils per packet.
=> $\dfrac{\frac{X}{n}*c}{T} = 1.2$
=> $T = \dfrac{\frac{X}{n}*c}{1.2}$ ———-(1)
If David packs $n-1$ pencils per packet, he will earn $20$% more profit. Therefore, the new profit will be 1.2P = 1.2*20 = $24$%.
=> $\dfrac{\frac{X}{n-1}*c}{T} = 1.24$
=> $T = \dfrac{\frac{X}{n-1}*c}{1.24}$ ——–(2)
Equating (1) and (2), we get,
$\dfrac{\frac{X}{n}*c}{1.2} = \dfrac{\frac{X}{n-1}*c}{1.24}$
$1.2*n = 1.24*(n-1)$
$0.04n = 1.24$
=> $n = 31$
We know that the pencils can be grouped into packets of 31 or packets of 30. Therefore, the total number of pens should be a multiple of 30 and 31. Also, since the least possible value of the number of pens is asked, the answer will be the LCM of 30 and 31.
LCM of 30 and 31 is 930 and hence, 930 is the right answer.
Question 70: A, B and C started working together on a project, but A left the work 9 days before the completion and B left the work 7 days before completion. Had A worked for one more day, the work would have been completed two days before the actual time. Had B worked for two days more, the work would have been completed three days before the actual time. In how many days can A, B and C together do a different work which can be done by C alone in 18 days?
70) Answer: 4
Solution:
If A works for one more day, the total number of days decreases by two. This means that C’s working days reduce by two. Thus, A is twice as efficient as C.
Similarly, if B works for two more days, C’s working days reduce by three.
Thus, B is 1.5 times more efficient than C.
The efficiencies of A, B and C are in the ratio 4 : 3 : 2.
New work is done by C in 18 days
=> New work = 2 * 18 = 36 units
(A + B + C)’s 1 day’s work = (4 + 3 + 2) = 9 units
Number of days required by all of them to complete the work = $\frac{36}{9}$ = 4 days
Hence, the correct answer is 4 days.
Question 71: Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is $\frac{3}{2}$ times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is
71) Answer: 150
Solution:
Each interior angle in an n-sided polygon = $\frac{\left(n-2\right)180\ }{n}$
It is given that each interior angle of B is $\frac{3}{2}$ times each interior angle of A and b = 2a
$\frac{\left(b-2\right)180\ }{b}$ = $\ \frac{\ 3}{2}\times\ $ $\frac{\left(a-2\right)180\ }{a}$
$2\times\ \left(b-2\right)\times\ a\ =\ 3\times\ \left(a-2\right)\times\ b$
2(ab-2a) = 3(ab-2b)
ab-6b+4a=0
a*2a-12a+4a=0
$2a^2-8a=0$
a(2a-8) = 0
a cannot be zero so 2a=8
a=4, b = 4*2=8
a+b = 12
Each interior angle of a regular polygon with 12 sides = $\ \frac{\ \left(12-2\right)\times\ 180}{12}$
=150
Question 72: What is the largest positive integer n such that $\frac{n^2 + 7n + 12}{n^2 – n – 12}$ is also a positive integer?
a) 6
b) 16
c) 8
d) 12
72) Answer (D)
Solution:
$\ \frac{\ n^2+3n+4n+12}{n^2-4n+3n-12}$
=$\ \frac{\ n^{ }\left(n+3\right)+4\left(n+3\right)}{n^{ }\left(n-4\right)+3\left(n-4\right)}$
=$\ \frac{\left(\ n+4\right)\left(n+3\right)}{\left(n-4\right)\left(n+3\right)}$
=$\ \frac{\left(\ n+4\right)}{\left(n-4\right)}$
=$\ \frac{\left(\ n-4\right)+8}{\left(n-4\right)}$
=$\ 1+\frac{8}{\left(n-4\right)}$ which will be maximum when n-4 =8
n=12
D is the correct answer.
Question 73: A total of 100 people attended a wedding. All the women who attended the wedding were married and were accompanied by their husbands. Each man, woman and a child gave Rs. 5, Rs. 4 and Rs. 1 as gift to the married couple. The married couple in total got Rs. 200 from the people who attended the wedding. It is known that at least half of the men who attended the wedding came along with their wives. What is the difference between the number of men and women who attended the wedding?
73) Answer: 4
Solution:
Let the number of men, women and boys who attended the wedding be m, w and c respectively.
It is known that,
m+w+c = 100 —– (1)
5m+4w+c = 200 —– (2)
4*(1)-(2) we get,
m = 3c-200 —– (3)
5*(1) – (2) we get,
w = 300-4c —– (4)
It is given that every woman came with her husband and at least half of the men came with their wives. So w must be between m/2 and m.
If w = m, from equations (3) and (4) we get,
c = 71.42
If w = m/2, from equations (3) and (4) we get,
c = 72.72
Since c must be an integral value it must be 72.
So from equations (3) and (4) we get,
m-w = (3c-200)-(300-4c) = 504-500 = 4
Question 74: One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be
a) 20200
b) 20500
c) 20800
d) 20000
74) Answer (B)
Solution:
Profit per boarder = Total profit / Number of boarders.
Let the number of boarders be n.
Profit/boarder = 1600 – (Total cost/n)
Let the total cost be a + bn, where a = fixed, and b is the variable additional cost per boarder.
Profit/boarder = 1600 – (a + bn)/n
Profit/boarder = 1600 – a/n – b
1600 – a/50 – b = 200
1600 – a/75 – b = 250
Solving, we get a = 7500, and b = 1250
Hence, total profit with 80 people = 80 ( 1600 – 7500/80 – 1250) = 80 (350 – 7500/80) = 28000 – 7500 = Rs. 20500
Question 75: If x is a real number such that $\log_{3}5= \log_{5}(2 + x)$, then which of the following is true?
a) 0 < x < 3
b) 23 < x < 30
c) x > 30
d) 3 < x < 23
75) Answer (D)
Solution:
$1 < \log_{3}5 < 2$
=> $ 1 < \log_{5}(2 + x) < 2 $
=> $ 5 < 2 + x < 25$
=> $ 3 < x < 23$
Question 76: Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is
a) $\frac{1}{\surd2}$
b) $\frac{\pi}{3}$
c) $\surd2$
d) 1
76) Answer (D)
Solution:
Let ‘h’ be the height of the triangle ABC, semiperimeter(S) $= \frac{4+4+r+4+4+r}{2} = 8+r$,
$a=4+r, b=4+r, c=8$
Area of triangle ABC $=\ \ \sqrt{\ s\cdot\left(s-a\right)\left(s-b\right)\left(s-c\right)}=$
$= \sqrt{\left(\ 8+r\right)\times\ 4\times\ 4\times\ r}$ = $\ \frac{\ 1}{2}\times\ \left(4+4\right)\times\ height$
Height (h) = $\sqrt{\ \left(8+r\right)r}$
Now, $ h + r = 4 \longrightarrow \sqrt{\ \left(8+r\right)r} + r = 4$ (Considering the height of the triangle)
$\sqrt{\ \left(8+r\right)r}$=4-r
16r=16
r=1
Alternatively,
$\text{AE}^@+\text{EC}^2=\text{AC}^2 \longrightarrow 4^2+\left(4-r\right)^2 = \left(4+r\right)^2 \longrightarrow\ \longrightarrow\ \longrightarrow\ r=1$
Question 77: If $x$, $y$ and $z$ are positive real numbers such that $x^{log_{4}{7}} = 64$, $y^{log_{9}{11}} = 81$, and $z^{log_{17}{25}} = \sqrt{17}$, then find the value of $x^{(log_{4}{7})^2}$+$y^{(log_{9}{11})^2}$+$z^{(log_{17}{25})^2}$.
77) Answer: 469
Solution:
$a^{(log_{b}{c})^2}$ = $a^{({log_{b}{c}*log_{b}{c}})}$ = $(a^{log_{b}{c}})^{log_{b}{c}}$
Therefore, we can write $x^{(log_{4}{7})^2}$ = $(x^{log_{4}{7}})^{log_{4}{7}}$.
It is given that $x^{log_{4}{7}} = 64$
Hence, $x^{(log_{4}{7})^2}$ = $(x^{log_{4}{7}})^{log_{4}{7}}$ = $(64)^{log_{4}{7}}$ = $(4)^{log_{4}{7^3}} = 343$
Similarly, $y^{(log_{9}{11})^2}$ = $(81)^{log_{9}{11}}$ = $(9)^{log_{9}{11^2}} = 121$
and $z^{(log_{17}{25})^2}$ = $(\sqrt{17})^{log_{17}{25}}$ = $(17)^{log_{17}{\sqrt{25}}} = 5$
Therefore, $x^{(log_{4}{7})^2}$+$y^{(log_{9}{11})^2}$+$z^{(log_{17}{25})^2} = 343+121+5 = 469$
Question 78: In a right triangle ABC, B is the right angle. A circle with centre O located on AB is drawn such that AC and BC are its tangents. If AO:BO = 25:24 and the perimeter is 392 cm, then find the length of AC(cm).
a) 175
b) 168
c) 182
d) 210
78) Answer (A)
Solution:
From the figure,
Assume, AO = 25x and BO = 24x
Since, AC is a tangent, ADO is a right angle.
So AD = $\sqrt{\left(25x\right)^2-\left(24x\right)^2}$ = 7x
Now in triangle ADO and ABC, $\angle\ AOD\ \ =\ \angle\ ACB$, $\angle\ ADO\ \ =\ \angle\ ABC$ = $90^{\circ\ }$
Hence, both the triangles are similar.
=> $\ \frac{OD\ }{AD}=\ \frac{\ BC}{AB}$
=> $\ \frac{24\ }{7}=\ \frac{\ BC}{49x}$
=> BC = 168x
Now, AC = $\sqrt{\ \left(168x\right)^2+\left(49x\right)^2}$ = $7\left(\sqrt{\ \left(24x\right)^2+\left(7x\right)^2}\right)$ = 175x
The perimeter = 49x+175x+168x = 392x = 392 => x=1,
Hence, AC= 175
Question 79: Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?
a) 45
b) 36
c) 32
d) 40
79) Answer (C)
Solution:
Let the efficiency of humans be ‘h’ and the efficiency of robots be ‘r’.
In the first case,
Total work = (15h + 5r) * 30……(i)
In the second case,
Total work = (5h + 15r) * 60……(ii)
On equating (i) and (ii), we get
(15h + 5r) * 30 = (5h + 15r) * 60
Or, 15h + 5r = 10h + 30r
Or, 5h = 25r
Or, h = 5r
Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h
Time taken by 15 humans = $\dfrac{\text{480h}}{\text{15h}}$ days= 32 days.
Hence, option C is the correct answer.
Question 80: Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250?
a) 6350
b) 6000
c) 7000
d) 7250
80) Answer (D)
Solution:
Assuming the investment of Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x
Given, Bina’s interest income exceeds Amala by 20x-18x=2x=250 => x=125
Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250
Question 81: A tank has two taps attached to it. Tap A fills the empty tank in 10 hours and tap B empties the full tank in 12 hours. At time t = 0, the tank is empty and both the taps are open. Tap A is closed at t = 30 min, reopened at t = 60 min, closed again at t = 90 min and so on. Tap B is closed at t = 45 min, reopened at t = 90 min, closed at t = 135 min and so on. Both the taps are closed as soon as the tank is full. If the tank is full at t = x minutes. Find x.
a) 115.5 hours
b) 120 hours
c) 118 hours 22.5 minutes
d) 118.5 hours
81) Answer (A)
Solution:
Tap A takes 10 hours to fill an empty tank and tap B takes 12 hours to drain a full tank. So, the efficiencies of A and B are in the ratio 12:10. So, A does 12 units of work in 1 hour and B does 10 units of work in 1 hour.
The duration of one cycle (open-close-open) for tap A is 60 minutes and the duration of one cycle (open-close-open) for tap B is 90 minutes.
LCM (60, 90) = 180
Consider the time from t = 0 to t = 180 min:
Tap A is open for 90 minutes. So, it does a work of 1.5*12 = 18 units.
Tap B is open for 90 minutes. So, it does a work of 1.5*10 = 15 units.
So, effective work done by both the taps in 180 minutes = 3 units.
Total work needed to be done = 10*12 = 120 units.
So, time required = 120/3 * 180 minutes = 40 * 3 hours = 120 hours.
But, the tank will be full before 120 hours.
Consider the first cycle: Work done in the first half an hour = 1 unit. In every other cycle, work done in the next 15 minutes is -2.5 units by tap B. But, in the first cycle, there is only 1 unit to drain out in the next 15 minutes. So, in effect, there is a gain of 1.5 units of work in the first cycle compared to the other cycles. That is, 4.5 units of work is done in the first cycle.
At the end of the 38th cycle, 115.5 units of work is done. Now, in the 39th cycle, an effective work of 4.5 units is done in the first 90 minutes. This is when 120 units of work is done and the tank is full. Both the taps are now closed.
So, total time required to fill the empty tank = (38*3 + 1.5) hours = 115.5 hours
Question 82: Find the number of ordered pairs $(a, b)$, such that $|a . b| = |a| + |b| + 72$. Given that $a$ and $b$ are integers.
82) Answer: 8
Solution:
$|a . b| = |a| + |b| + 72$
$|a| * |b| – |a| – |b| = 72$
$|a|(|b| – 1) – (|b| – 1) – 1 = 72$
$(|a| – 1)(|b| – 1) = 73$
The only possible solutions are (|a|, |b|) = (2, 74) or (74, 2)
Each solution has 4 solutions for (a, b).
Hence, there are a total of 8 possible solutions.
Question 83: AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
a) 9.3
b) 7.8
c) 9.1
d) 8.5
83) Answer (C)
Solution:
Since AB is a diameter, AQB and APB will right angles.
In right triangle APB, AP = $\sqrt{10^2-6^2}=8$
Now, 2AQ=AP => AQ= 8/2=4
In right triangle AQB, AP = $\sqrt{10^2-4^2}=9.165$ =9.1 (Approx)
Question 84: The number of solutions $(x, y, z)$ to the equation $x – y – z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is
a) 101
b) 99
c) 87
d) 105
84) Answer (B)
Solution:
x – y – z = 25 and $x\leq40,y\leq12$, $z\leq12$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99
Question 85: The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?
a) 3.5
b) 5
c) 4.5
d) 4
85) Answer (C)
Solution:
It is given that the average of the 30 integers = 5
Sum of the 30 integers = 30*5=150
There are exactly 20 integers whose value is less than 5.
To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers
So the sum of 10 integers = 10*6=60
The sum of the 20 integers = 150-60= 90
Average of 20 integers = $\ \frac{\ 90}{20}$ = 4.5
Question 86: A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?
a) 4:48 pm
b) 4:12 pm
c) 4:24 pm
d) 4:36 pm
86) Answer (C)
Solution:
Let ‘t’ pm be the time when the tank is emptied everyday. Let ‘a’ and ‘b’ be the liters/hr filled by pump A and pump B respectively.
On Monday, A alone completed filling the tank at 8 pm. Therefore, we can say that pump A worked for (8 – t) hours. Hence, the volume of the tank = a*(8 – t) liters.
Similarly, on Tuesday, B alone completed filling the tank at 6 pm. Therefore, we can say that pump B worked for (6 – t) hours. Hence, the volume of the tank = b*(6 – t) liters.
On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 – t) hours and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 – t)+2b liters.
We can say that a*(8 – t) = b*(6 – t) = a*(5 – t) + 2b
a*(8 – t) = a*(5 – t) + 2b
$\Rightarrow$ 3a = 2b … (1)
a*(8 – t) = b*(6 – t)
Using equation (1), we can say that
$a*(8-t)=\dfrac{3a}{2}*(6-t)$
$t = 2$
Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone.
Hence, working together both can fill the tank in = \dfrac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes.
The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.
Question 87: Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is
a) 1680
b) 1176
c) 2520
d) 1440
87) Answer (A)
Solution:
It is given,
7C = 30P = 9A and Ankita bought 4C, 14P and 6A.
Let 7C = 30P = 9A = 630k
C = 90k, P = 21k, and A = 70k
Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k
Marked up price = 1074k + 1752
S.P = $\frac{1}{6}\left(1074k+1752\right)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)\left(1074k+1752\right)$ = $\frac{5}{6}\left(1074k+1752\right)$
S.P – C.P = profit
$1460-\frac{1074k}{6}=744$
$\frac{1074k}{6}=716$
k = 4
Money spent on buying almonds = 420k = 420*4 = Rs 1680
The answer is option A.
Question 88: Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is
a) 20
b) 30
c) 25
d) 10
88) Answer (A)
Solution:
Let the speed of cars be a and b and the distance =d
Minimum time taken by 1st car = 6 hours,
For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM.
Hence, car 2 will take 5 hours.
Hence a= $\ \frac{\ d}{6}$ and b = $\ \frac{\ d}{5}$
Hence the speed of car 2 will exceed the speed of car 1 by $\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$ = $\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$ = 20
Question 89: In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is
89) Answer: 1098
Solution:
In a tournament, there are 43 junior level and 51 senior level participants.
Let ‘n’ be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.
$\Rightarrow$ nC2 = 153
$\Rightarrow$ n(n-1)/2 = 153
$\Rightarrow$ n(n-1) = 306
=> n$^{2}$-n-306 = 0
=> (n+17)(n-18)=0
=> n=18 (rejecting n=-17)
Therefore, number of boys on junior level = 43 – 18 = 25.
Let ‘m’ be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.
$\Rightarrow$ mC2 = 276
$\Rightarrow$ m = 24
Therefore, number of girls on senior level = 51 – 24 = 27.
Hence, the number of matches a boy plays against a girl = 18*25+24*27 = 1098
Question 90: Suppose k is any integer such that the equation $2x^{2}+kx+5=0$ has no real roots and the equation $x^{2}+(k-5)x+1=0$ has two distinct real roots for x. Then, the number of possible values of k is
a) 9
b) 7
c) 8
d) 13
90) Answer (A)
Solution:
$2x^{2}+kx+5=0$ has no real roots so D<0
$k^2-40\ <0$
$\left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)<0$
$k\in\left(-\sqrt{40},\sqrt{40}\right)$
$x^{2}+(k-5)x+1=0$ has two distinct real roots so D>0
$\left(k-5\right)^2-4>0$
$k^2-10k+21>0$
$\left(k-3\right)\left(k-7\right)>0$
$k\in\left(-\infty\ ,3\right)∪\left(7,\infty\ \right)$
Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2
In 9 total 9 integer values of k are possible.
Question 91: A company publishes to its customers that at a certain compound interest rate, a sum of money deposited by anyone will become 8 times in 3 years. If the same amount is deposited at the same compound rate of interest, then in how many years will it become 16 times ?
a) 4 years
b) 3.5 years
c) 3 years
d) 4.5 years
91) Answer (A)
Solution:
If a sum of money becomes $x$ times in $a$ years, it will become $y$ times in $b$ years
=> $(x)^{\frac{1}{a}}=(y)^{\frac{1}{b}}$
Similarly, $(8)^{\frac{1}{3}}=(16)^{\frac{1}{t}}$
=> $(2)^{\frac{3}{3}}=(2)^{\frac{4}{t}}$
=> $1=\frac{4}{t}$
=> $t=4$ years
=> Ans – (A)
Question 92: If $(5.55)^x = (0.555)^y = 1000$, then the value of $\frac{1}{x} – \frac{1}{y}$ is
a) $\frac{1}{3}$
b) 3
c) 1
d) $\frac{2}{3}$
92) Answer (A)
Solution:
We have, $(5.55)^x = (0.555)^y = 1000$
Taking log in base 10 on both sides,
x($\log_{10}555$-2) = y($\log_{10}555$-3) = 3
Then, x($\log_{10}555$-2) = 3…..(1)
y($\log_{10}555$-3) = 3 …..(2)
From (1) and (2)
=> $\log_{10}555$=$\ \frac{\ 3}{x}$+2=$\ \frac{\ 3}{y}+3$
=> $\frac{1}{x} – \frac{1}{y}$ = $\frac{1}{3}$
Question 93: Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is
a) 10
b) 5
c) $8\surd2$
d) $6\surd2$
93) Answer (A)
Solution:
Let p be the length of AP.
It is given that $\angle\ BAC\ =90\ $ and $\angle\ APC\ =90\ $
Let $\angle\ ABC\ =\theta\ $, then $\angle\ BAP\ =90-\theta\ $ and $\angle\ BCA\ =90-\theta\ $
So $\angle\ PAC\ =\theta\ $
Triangles BPA and APC are similar
$p^2=x\left(20-x\right)$
We have to maximize the value of p, which will be maximum when x=20-x
x=10
Question 94: ABCD is a rectangle as shown in the figure. AB = 8 cm and BC = 6 cm. BE is the perpendicular drawn from B to the diagonal AC. EF is the perpendicular drawn from E to AB. What is the length of BF?
a) 3.24 cm
b) 1.96 cm
c) 2.56 cm
d) 2.88 cm
94) Answer (D)
Solution:
By Pythagoras theorem $AC=\sqrt{AB^2+BC^2}=\sqrt{8^2+6^2}=10$ cm
Triangle $BEC$ is similar to triangle $ABC$. So we get,
$\frac{BE}{AB}=\frac{BC}{AC}$
=> $\frac{BE}{8}=\frac{6}{10}$
=> $BE=4.8$ cm
Triangle $BFE$ is similar to triangle $BEA$. So we get,
$\frac{BF}{BE}=\frac{BE}{AB}$
=> $BF=\frac{BE^2}{AB}$
=> $BF=\frac{4.8^2}{8}$
=> $BF=2.88$ cm
Question 95: Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is
a) 6
b) 15
c) 10
d) 12
95) Answer (B)
Solution:
M – First meeting point
Let the speeds of trains A and B be ‘a’ and ‘b’, respectively.
$\frac{x}{a}=\ \frac{\ D-x}{b}$
It is given,
$\frac{D}{a}=10$ and $\frac{x}{b}=9$
$\frac{x}{\frac{D}{10}}=\ \frac{\ D-x}{\frac{x}{9}}$
$\frac{10x}{D}=\ \frac{\ 9D-9x}{x}$
$10x^2=\ \ 9D^2-9Dx$
$10x^2+9Dx-9D^2=\ 0$
Solving, we get $x=\frac{3D}{5}$
$\frac{x}{b}=9$
$\frac{3D}{b\times5}=9$
$\frac{D}{b}=15$
The total time taken by train B to travel from station Y to station X is 15 minutes.
The answer is option B
Question 96: A solid cylindrical wafer of height 10 cms and radius 3.5 cms is cut into two equal halves along its height. The two halves are then attached to each other along their flat surfaces such that they make a cross.(90 degree angle with each other). If the cross is then coated uniformly with 1mm layer of chocolate, what is the volume of chocolate needed? (in $cm^3$)
a) 20.25
b) 33.9
c) 34.25
d) 35.3
96) Answer (B)
Solution:
Look at the diagram given below
Here, AD will be 7 cms and EF will be 10 cms. So, the cross section formed will be a square of side 7 cms.
The area can be calculated as taking the total surface area of the cylinder+The rectangles formed when the cylinder is cut into two-the cross section when they are stuck together.
Since the wafers make a 90 degree angle with each other, the area is a square of side 7 cms.
=$2\pi*r*h+2\pi(r)^2+7(10)(2)-2*7^2$
=339 $cm^2$
Volume of chocolate needed = 339*0.1
=33.9 $cm^3$
Question 97: How many ten-digit numbers can be formed using all the digits of 2435753228 such that odd digits appear only in even places?
a) $2!3!5!$
b) $(5!)^{2}$
c) $\frac{(5!)^{2}}{3!}$
d) $\frac{(5!)^{2}}{3!(2!)^{2}}$
97) Answer (D)
Solution:
we have : 2435753228
Now here we have 5 odd digits and 5 even digits and total 10 places
so 5,5,3,3,7 will arrange in 5 places and 2,2,2,8,4 will arrange in 5 places
we get : total arrangements =$\frac{5!}{3!}\times\ \frac{5!}{2!\times\ 2!}$
we get $\frac{\left(5!\right)^2}{3!\ \left(2!\right)^2}$
Question 98: Rajesh, a courier delivery agent, starts at point A and makes a delivery each at points B, C and D, in that order. He travels in a straight line between any two consecutive points. The following are known: (i) AB and CD intersect at a right angle at E, and (ii) BC, CE and ED are respectively 1.3 km, 0.5 km and 2.5 km long. If AD is parallel to BC, then what is the total distance (in km) that Rajesh covers in travelling from A to D?
a) 10.2
b) 12
c) 11.5
d) 5.5
e) 18
98) Answer (C)
Solution:
Given, CE=0.5, BC = 1.3 and ED=2.5
Triangle CEB is a right-angled triangle => EB = 1.2
Triangles ECB is similar to triangle EDA
EB/EC = AE/ED => AE = 6
Hence total distance travelled = AB + BC + CD = 7.2 + 1.3 + 3.5 = 11.5km
Question 99: The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is
a) 3 : 10
b) 1 : 3
c) 1 : 4
d) 2 : 5
99) Answer (B)
Solution:
Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.
It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.
$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$
$\Rightarrow$ $a+2b+3c = 120$ … (1)
If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.
$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$
$\Rightarrow$ $3a+2b+c = 180$ … (2)
From equation (1) and (2), we can say that
$\Rightarrow$ $b+2c = 45$
$\Rightarrow$ $b = 45 – 2c$
Also, on subtracting (1) from (2), we get
$a – c = 30$
$\Rightarrow$ $a = 30 + c$
In solution D, B and C are mixed in the ratio 2 : 7
So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ = $\dfrac{90 – 4c + 7c}{9}$ = $\dfrac{90 + 3c}{9}$
Required ratio = $\dfrac{90 + 3c}{9a}$ = $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$
Hence, option B is the correct answer.
Question 100: An article is marked x% above the cost price. A discount of $\frac{2}{3}$x% is given on the marked price. If the profit is 4% of the cost price and the value of x lies between 25 and 50, then the value of 50% of x is?
a) 12
b) 16
c) 13
d) 15
e) 20
100) Answer (D)
Solution:
Let CP of object be a.
It is given that SP = $(1+\frac{4}{100} )$ x CP
$ \Rightarrow $ SP = 1.04 x CP
It is given that MP = $(1+\frac{x}{100} )$ x CP
It is also given that SP = $(1-\frac{\frac {2x}{3}}{100} )$ x MP
$ \Rightarrow $ SP = $(1-\frac{2x}{300} )$ x $(1+\frac{x}{100} )$ x CP
$ \Rightarrow (1+\frac{4}{100})$ x CP = $(1-\frac{2x}{300} )$ x $(1+\frac{x}{100} )$ x CP
$ \Rightarrow (1+\frac{4}{100})$ = $(1-\frac{2x}{300} )$ x $(1+\frac{x}{100} )$
$ \Rightarrow \frac{104}{100}$ = $(1-\frac{2x}{300} )$ x $(1+\frac{x}{100} )$
$ \Rightarrow \frac{104}{100}$ = $(\frac{300-2x}{300} )$ x $(\frac{x+100}{100} )$
$ \Rightarrow \frac{104}{100} \times 300 \times 100$ = $(300-2x)$ x $(x+100)$
$ \Rightarrow 31200 $ = $(300-2x)$ x $(x+100)$
Now, we look at the options.
Since the question says that $x\epsilon[25,30]$ so 50% of x ie 0.5x cannot be less than 12.50 ie $\frac {25}{2}$ and cannot be more than 25 ie $\frac {50}{2}$
This eliminates option A.
Putting values of $x$ given in the remaining options in the final expression, [here we need to be careful to use the value of x and not the value of 50% of x as given in the expression]
Look carefully in the remaining options : 16,13,15 as 0.5x ie 32,26,30 as possible values of x.
In the question since the discount rate offered is $\frac{2x}{3}$%, then a safer choice would be to check for the option of 30 in the beginning. It is a safer choice because the percentage of discount that we get in the other options are not whole numbers.
NOTE : THIS IS JUST A SAFE CHOICE AND NEVER MARK AN ANSWER DIRECTLY ON THIS PRESUMPTION WITHOUT CHECKING IT.
Putting x=30 in the expression :
(300 – (2×30))x(30+100) = (300-60)x(100+30) = 240×130 = 24x13x100= 312×100 = 31200= LHS of expression.
Thus the value of $x$ is 30.
Therefore value of 50% of $x$ = 0.5×30 = 15
