If x, y, z are three positive numbers such that (A - x): (A - y): (A - z) = 1: 7: 4 and 2A = x+y+z, then x:y:z=
AP Police SI Mains 2019 Arithmetic and Reasoning
For the following questions answer them individually
Solution
Let say,
$$A-x=k\ ,\ A-y=7k\ and\ A-z=4k.$$
So,$$2A=2\left(k+x\right)\ ,\ 2A=2\left(7k+y\right)\ and\ 2A=2\left(4k+z\right).$$
So,$$x+y+z=2k+2x\ ,\ x+y+z=14k+2y\ and\ \ x+y+z=8k+2z.$$.........................................(1)
or,$$3\left(x+y+z\right)=2\left(x+y+z\right)+2k+14k+8k.\ \ \left(by\ adding\ these\ 3\ equations\right)$$
or,$$x+y+z=24k.$$
put this value in (1) :
$$2x=22k\ or\ x=11k.$$
$$2y=10k\ or\ y=5k.$$
and $$2z=16k\ or\ z=8k.$$
So,$$x:y:z=11:5:8.$$
C is correct choice.
If x ≠ y ≠ z ≠ 0 , $$a^x = b^y = c^z$$ and $$\frac{a}{b} = \frac{b}{c}$$, then $$\frac{y - x}{z - y}$$ =
Solution
Let say,
$$a^x=b^y=c^z=k.$$
or,$$a=k^{\frac{1}{x}},\ b=k^{\frac{1}{y}}\ and\ c=k^{\frac{1}{z}}.$$
and $$\frac{a}{b}=\frac{b}{c}.$$
or,$$b^2=ac.$$
So,$$k^{\frac{2}{y}}=k^{\frac{1}{x}}k^{\frac{1}{y}}.$$
or,$$k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}.$$
or,$$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}.$$
or,$$2xz=yz+xy.\ \left(multiply\ both\ side\ by\ xyz\right)$$
or,$$yz-xz=xz-xy.$$
or,$$z\left(y-x\right)=x\left(z-y\right).$$
or,$$\frac{\left(y-x\right)}{\left(z-y\right)}=\frac{x}{z}.$$
So, A is correct choice.
Two different discounts x% and y% are allowed on two items having same cost price and marked price. If $$P_1$$% and $$P_2$$% are respectively the profits on them, x - y =20 and $$P_2 - P_1$$ = 32, then the ratio of their cost price to their marked price is
Solution
Let say, cost price=c and marked price=m.
given that , x-y=20.
No, SP_1=m(1-x) and SP_2=m(1-y) .
So,P1 %= (SP_1/c)-1=((m/c)(1-x))-1.
and P2 %=(SP_2/c)-1=((m/c)(1-y))-1.
so,( P2-P1)
=((m/c)(1-y))-1-((m/c)(1-x))+1
=(m/c)(1-y-1+x)
=(m/c)(x-y).
=20m/c.
So,
20m/c=32.(given in question)
or,c/m=20/32
or,c/m=5/8.
So,c:m=5:8.
C is correct choice.
If 0 < a < b then, for all x > 0, $$\frac{a + x}{b + x}$$ >
Solution
Let assume any value for a,b and x that satisfy the above conditions.
Let say, a=1 and b=2 and x=5.
Then (a/b)=(1/2)=0.5.
and (a+x)/(b+x)=6/7=0.85.
So,(a+x)/(b+x) >(a/b).
but it is not greater than x or (b/a).
now,(1/x)=1/5=0.2
So,in this case (a+x)/(b+x)>(1/x).
But if we consider x=0.5 :
(a+x)/(b+x)=1.5/2.5=0.6.
but (1/x)=10/5=2.
So, in this case (a+x)/(b+x) is lesser than (1/x).
So, only option C is correct.
Four students A, B, C and D are running around a circular playground in a college and they take 1. 5 min, 25min, 35min and 20min respectively to complete one round. If a prize money of Rs. K is divided in the ratio of their speeds, then the student who receives_maximum share is
Solution
Let say, distance cover by them in the playground is d.
So, Speed of A,B,C and D are
(d/1.5),(d/25),(d/35) and (d/20).
So,they will receive money in following ratio:
(d/1.5):(d/25):(d/35):(d/20)
=(1/1.5):(1/25):(1/35):(1/20)
=(2100/1.5):(2100/25):(2100/35):(2100/20)
=1400:84:60:105.
So,
A will get=1400K/1649.
B will get=84K/1649.
C will get=60K/1649.
D will get=105K/1649.
So, A will get maximum amount.
A is correct choice.
The ratio among the following that will have maximum change in its value when 10 is added to both the antecedent and consequent of that ratio, is
Solution
According to given condition:
(15/17)-(5/7)=0.1680.
(12/13)-(2/3)=0.2564.
(14/17)-(4/7)=0.2521.
(13/14)-(3/4)=0.1785.
So, B is correct choice.
Distinct number of men, women and children have visited a park in a particular day. The number ofwomen visitors is 7 and the number of children who visited the park is maximum among all the visitors. If the ratio between the number of men and women is the same as the ratio between the number of women and children, then the total number of visitors to the park is
Solution
Number of men ,women and children are distinct number (given).
And number of women is 7.
Again,
number of men/number of women =
number of women/number of children.
So,number of men×number of children
=$$(number of women)^2$$
=$$7^2$$
=49.
But, 49 has three different factors 1,7 and 49.
as number of men, women and children are distinct, number of children or men can have either 49 or 1 as their value.
So,Total number of visitors were
=49+7+1=57.
C is correct choice.
If a person covers the same distance by walking, by cycling and by running with speeds in the ratio 1: 9: 3 respectively, then the ratio of the ratios of their times and speeds of the three modes of travel is
Solution
Let say ,speed of walking, cycling and running are k,9k and 3k respectively.
Time and speed are inversely proportional when distance is constant.
So,their time taken would be as following:
(1/k),(1/9k) and (1/3k).
So,ratio of time and speed would be :
(1/k)/k ,(1/9k)/9k and (1/3k)/3k
or,$$(1/k^2), (1/81k^2) and (1/9k^2)$$.
So , required ratio would be ==>
$$(1/k^2):(1/81k^2):(1/9k^2)$$
=$$1:(1/81):(1/9)$$
=81:1:9.
C is correct choice.
If the sum of the ratios equivalent to $$16\frac{2}{3}$$%, $$33\frac{1}{3}$$%, 5% and 25% is x% of 15, then x =
Solution
$$16\frac{2}{3}$$%=50/300.
$$33\frac{1}{3}$$%=100/300.
5%=5/100=15/300.
25%=25/100=75/300.
So, according to question:
(15/100)x=(50/300)+(100/300)+(15/300)+ (75/300).
(15/100)x=(240/300).
or,x=(240×100)/(15×300)=80/15=16/3.
A is correct choice.
A student is asked to multiply a number K by 5. But he multiplies K by $$\frac{1}{5}$$ and gets x% error. Next time he multiplies K by $$\frac{1}{25}$$ and gets y% error. Then the percentage increase of y over x is
Solution
Originally expected 5K.
but he wrote (1/5)K=0.2K.
first error,x%=(5K-0.2K)/5K=0.96 or 96%.
next time he wrote (1/25)K=0.04K.
second error,y%=(5K0.04K)/5K=0.992 or 99.2%.
so increase of y over x is =
(0.992-0.96)/0.96=0.033333 or 3.33%.
A is correct choice.
If there is an increase of 25% and 4% successively in the sales of a car of brand X in two consecutive years, then the net increase in the sales after two years, is
Solution
Let say ,initial number of cars were x.
So,after two successive increment in sales of 25% and 4% would be=x×1.25×1.04=1.3x.
So,net increment would be ((1.3x-x)/x)×100
=30%.
D is correct choice.
A variable x is proportional to y. If 3 values $$x_1, x_2, x_3$$ of x are in the ratio 2:3:4 such that $$x_1 + x_2 + x_3$$ = 9 and $$x_1y_1+x_2y_2+x_3y_3$$ = 29 then the ratio of the increase percentages of $$x_1y_1$$, $$x_2y_2$$, $$x_3y_3$$ over $$x_1,x_2,x_3 $$respectively is
Solution
$$x_1, x_2, x_3$$ of x are in the ratio 2:3:4.
Let say,$$x_1=2k,x_2=3k,x_3=4k$$
So,2k+3k+4k=9
or,k=1.
So,$$x_1=2,x_2=3,x_3=4$$
Now,
$$x_1y_1+x_2y_2+x_3y_3$$ = 29
or,$$2y_1+3y_2+4y_3$$ = 29.
By using trial and error, we can say that
$$y_1=2,y_2=3 and y_3=4$$ can be the values.
so,$$x_1y_1=4,x_2y_2=9,x_3y_3=16$$.
So, required increments are 100%,200% and 300%.
So, required ratio is 100:200:300=1:2:3.
C is correct choice.
A fruit vendor has certain oranges with him. He sells each orange for Rs.5. Three customers A, B, C successively bought 25%, $$33\frac{1}{3}$$, 50% of the oranges that are left over with the vendor each time. Later a fourth customer D bought 4 oranges. If A and D together paid Rs.140/- to the vendor, then the percentage of oranges left with the vendor is
Solution
Let say ,he had x number of oranges initially.
A bought 0.25x.
he left with 0.75x.
B bought (1/3)0.75x.
he left with (2/3)0.75x.
C bought (1/2)(2/3)0.75x.
he left with (1/2)(2/3)0.75x.
D bought 4 oranges.
So, finally he left with ((1/2)(2/3)0.75x-4).
According to question :
(0.25x)×5+4×5=140.
or,x=96.
So,he will be left with=((1/2)(2/3)0.75×96-4)
=20.
So, percentage would be=20/96=20.833%.
C is correct choice.
A, B, C and D are four students in a class. A's total score is 20% less than B's total score, C's total score is 25% more than A's total score, D's total score is 20% more than A's total score. If the least total score among the scores is 240, then the ratio of the scores of the four students in the decreasing order is
Solution
Let say A,B,C,D' scores are a,b,c,d respectively.
According to question,
a=0.80b , c=1.25a and d=1.2a
So, we can say that:
c=1.25×0.80b=b
and d=1.2a×0.80b=0.96b.
So,lowest among 0.80b,b,b and 0.96b is 0.80b.
So,0.80b=240
or,b=300.
So,c=300 and d=288.
So, required ratio
300:300:288:240
=25:25:24:20
C is correct choice.
A merchant is selling goods by importing from abroad. He gets a discount of $$33\frac{1}{3}$$% on 3 the catalogue price, pays 20% import duty on the net cost of the goods and sells the goods for a profit of 25%. If the catalogue price of an article is Rs. 3,756, then its selling price (in Rs.) is
Solution
Catalogue price 3756 Rs.
after discount price became =(2/3)×3756
=2504 Rs.
After import duty price became=1.2×2504
=3004.8 Rs.
Now he sells it for 25% profit.
So, selling price=1.25×3004.8=3756Rs.
D is correct choice.
A dishonest dealer claims to sell his goods for the cost price. If he uses 20% less weight in weighing the goods, his gain % is:
Solution
Let say ,he was selling 100 weight of products and cost price of the product c.
So,he was actually selling 0.8×100=80 weight of products.
80 unit were selling for c Rs.
1unit was selling for c/80 Rs.
So, 100 units were selling for (100/80)c
=1.25c Rs.
So, his overall gain was=((1.25c-c)/c)×100
=25%.
C is correct choice.
If a person sold his watch for Rs.24 with a profit percentage numerically equal to its cost price, then the cost price of the watch is Rs.
Solution
Let say cost price is c.
So, profit percentage=c%.
Selling price=c(1+c/100).
So,
c(1+c/100)=24.
or,$$c^2+100c-2400=0$$
or,$$c^2+120c-20c-2400=0$$
or,$$c(c+120)-20(c+120)=0$$
or,(c+120)(c-20)=0.
so,either c=-120 or c=20.
Cost price cannot be negative value.
so, c =20.
C is correct choice.
A shopkeeper offers successive discounts of 20% and 25% on the marked price of an article and gets a profit of 20%. If he wants to make 40% profit, the percentage by which the marked price is to be increased is
Solution
Let say,marked price is m and cost price is c.
So,after two successive discount ,he will sell it for = 0.8×0.75m=0.6m Rs.
Profit=(0.6m-c) Rs.
So, 0.6m-c=.2c
or,1.2c=0.6m
or, c/m=1/2.
Let say ,c=k and m=2k.
if he wants to profit 40% then selling price would become (1.4×k)=1.4k.
let say marked price should increase by n%.
then, (2k(1+n%)0.8×0.75-k)=0.4k.
or,1.2(1+n%)k=1.4k.
or,1+n%=14/12.
or,1+n%=7/6.
or,n%=1/6.
or,n=100/6=50/3.
C is correct choice.
A shopkeeper sells two types of articles A and B for the same price at Rs. 150/-. The cost prices of them are respectively Rs. 120/- and Rs. 200/-. On the first day he sells only one item of A and increases this number by 6 units each day. He sells 50 units of 13 on first day and decreases this number each day by 2 units. The number of days the shopkeeper incurs a net loss continuously is
Solution
Selling price of A and B are 150 Rs.
Cost price of A and B are 120 and 200 Rs.
profit on A=(150-120)=30 Rs.
and Profit on B=(150-200)= -50 Rs(- means loss)
1st day:
Total loss=30×1-50×50=-2470 Rs.
2nd day:
Total loss=30×7-50×48=-2190 Rs.
3rd day:
Total loss=30×13-50×46=-1910 Rs.
4th day:
Total loss=30×19-50×44=-1630 Rs.
5th day:
Total loss=30×25-50×42=-1350 Rs.
6yh day:
Total loss=30×31-50×40=-1070 Rs.
7th day:
Total loss=30×37-50×38=-790 Rs.
8th day:
Total loss=30×43-50×36=-510 Rs.
9th day:
Total loss=30×49-50×34=-230 Rs.
10th day:
Total profit=30×55-50×32=50 Rs.
So, loss will continue for 9 days.
D is correct choice.
The following table shows the different number of items a shopkeeper sold with different cost prices and different selling prices. Use this information to match the items of List A with the items of List B.
The correct match for i, ii, iii is
Solution
For
100 items. : Profit=(14-10)100=400.
200 items: profit=(12.5-9.5)200=600.
300 items: profit=(11-9)300=600.
400 items: profit=(10-8)400=800.
500 items: profit=(9-7.5)500=750.
600 items: profit=(7.5-7)600=300.
700 items: profit=(6.5-6)700=350.
So, profit for 800 items is maximum.
So, i will go with a.
if selling price is increased by 1 ,then profit for 100 items become=(15-10)100=500.
This is the minimum profit for the ii condition.
So,ii will go with b.
if no conditions are given then,
profit for 600 items will be minimum which incurs only 600(6.5-6)=300.
So, iii will go with c.
So, i,ii and iii will go with a,b and c.
A is correct choice.
The average score of 3 students A, B and C is 72. When D joins them the average score of all the four becomes 70. If another student E, whose score is 4 more than that of D replaces A then the average score of B, C, D and E becomes 68. Then the score of A
Solution
Total score of A,B,C is 72×3=216.
after adding D's score total become 70×4=280.
So,D's score=280-216=64.
E's score=64+4=68.
Now ,(A+B+C+D+E)=216+64+68=348.
But total of B,C,D,E is=68×4=272.
So,A's score=348-272=76.
D is correct choice.
The first quality of juice costs Rs.15 per litre and the second quality of juice costs Rs.10 per litre. If the mixture of these two qualities is sold at the rate of Rs.14 per litre, then the ratio in which these two qualities of juices are to be mixed in order to get a profit of 20% is
Solution
Let say, he mixed x of first quality with y of second quality.
Ao,price of x amount of first quality is 15x.
and price of y amount of second quality is 10y.
but he sold (x+y) in 14(x+y).
So,
(15x+10y)1.2=14(x+y)
or,18x+12y=14x+14y.
or,4x=2y.
or,2x=y.
So, x/y=1/2=(1/2)/1.
so,x:y=(1/2):1.
C is correct choice.
The average of all the numbers which are the first ten multiples of each of the first ten natural numbers is
Solution
10 multiples of
1:1,2,3,4,5,6,7,8,9,10
2:2,4,6,8,10,12,14,16,18,20
And so on but if you observe here,
If 2 is taken out from the series we get,2(1+2+3+4+5+6+7+8+9+10)
Therefore the series comes out to be
S=1(1+2+3+4+…..+10) +2(1+2+3+4+5+6+…+10)+3(1+2+3+4+5+6+…+10)+4(1+2+3+4+5+6+…+10)+5(1+2+3+4+5+6+…+10)+5(1+2+3+4+5+6+…+10)+…….+10(1+2+3+4+5+….+10)
S=(1+2+3+4+5+6+7+8+9+10)(1+2+3+4+5+6+7+8+9+10)
S=((10*11)/2)*((10*11)/2)
S=(110*110)/4
That is the sum comes out to be S
Average is S/100 since there are total 100 numbers
This implies
A=(110*110)/(4*100)
A=121/4=30.25.
A is correct choice.
The ratio of copper to zinc in an alloy 'A' of 7kgs is 5:2. The ratio of the same metals in that order in another alloy 'B' of 7kgs is 3:4. If 28kg of alloy is made by mixing A and B in quantities x & y respectively so as to have the ratio of copper and zink in the ratio 1:1, then x : y is
Solution
Copper and Zinc ratio in A and B are 5:2 and 3:4.
According to question,
x+y=28...................(1)
in x quantity of A copper and zinc are as following :
copper=(5x/7) and zinc =(2x/7).
in y quantity of B copper and zinc are as following :
copper=3y/7 and zinc=(4y/7).
So,total copper=(5x/7+3y/7)
and total zinc=(2x/7+4y/7).
According to question:
(5x/7+3y/7)=(2x/7+4y/7)
or,5x+3y=2x+4y.
or,3x=y. Now put this value in (1).
4x=28 or, x=7.
So,y=28-7=21.
So,x:y=7:21=1:3.
D is correct choice.
A milk vendor generally sells 3 Grades of milk. Grade I is pure milk with no water mixed in it, Grade II is a mixture of milk and water in the ratio 3:2 and Grade III is a mixture of milk and water in the ratio 2:3. On a particular day he has x liters of Grade I and 3 liters of Grade III milk and he got an order to supply 7 liters of Grade II milk. The minimum value ofx (in litres) required to prepare 7 Its of Grade H milk by mixing Grade I milk, Grade III milk and water, is
Solution
He needs to supply 7 ltr of Grade II milk : which contain milk:water in 3:2 ratio.
So, He must need $$\left(7\times\frac{3}{5}\right)=4.2$$ ltr of milk.
But ,if we consider only Grade III mixture ,then we have only $$\left(3\times\frac{2}{5}\right)=1.2$$ltr of milk available.
(as Grade III mixed in 2:3 ratio).
So, minimum amount of Grade I milk needed was (4.2-1.2)=3 ltr ,by which he can make
7 ltr of Grade II milk as well as Grade H milk mixtures by adding water to rest of the mixtures.
A is correct choice.
A business man buys two qualities A and B of a product at Rs. 120 per kg and Rs. 60 per kg respectively. He then mixes these two qualities and sells at Rs. 100 per kg. Then the percentage increase in the profit on a certain quantity of the mixture of A and B in the ratio 7:11 on the profit on the same quantity of the mixture of A and B in the ratio 1:1, is
Solution
Business man buys two qualities A and B of a product at Rs. 120 per kg and Rs. 60 per kg respectively.
Let say, When he mixes A and B in 1:1, he produces 2kg of 180 Rs. product.
So, 1kg of mixture cost price 90 Rs.
He sold it at 100 Rs/kg.
So,Profit=100-90=10 Rs/kg.
Again ,
Let say, When he mixes A and B in 7:11, he produces 18 kg of (840+660)=1500 Rs. product.
So, 1 kg of mixture cost price $$\frac{1500}{18}$$ Rs.
So, profit=$$100-\frac{1500}{18}=\frac{300}{18}=\frac{50}{3}$$ Rs/kg.
So, percentage increase in profit=$$\frac{\left(\frac{50}{3}-10\right)}{10}\times100=\frac{20}{30}\times100=\frac{200}{3}.$$
D is correct choice.
A jar contains a mixture of 2 liquids A and B in the ratio 4:1. If 10 liters of mixture is taken out and 10 liters of liquid B is poured into the jar, the ratio becomes 2:3. The amount of liquid A contained in the jar initially is
Solution
Let say, liquid A and B in the jar are 4k and k.
if 10 litter of mixture is taken ,then amount of A and B in the jar would be (10×4/5) and (10×1/5) or 8 litter and 2 litter respectively.
So,in the mixture A will be left 4k-8 and B will be left k-2.
Now if we add 10 litter of B ,then new amount of B will be=k-2+10=k+8.
So,
(4k-8)/(k+8)=2/3.
Or,12k-24=2k+16.
or,10k=40.
or,k=4.
So, initially amount of A was 4×4=16 litter.
D is correct choice.
The amount of water to be mixed with 32 liters of pure fruit juice so as to get 25% profit on selling the mixture at the cost price of the pure juice, (in liters) is
Solution
Let say ,CP of 32 ltr juice is 100 Rs and x ltr water to be mixed in the juice.
Then (32+x) ltr of juices selling price is$$=\frac{100\left(32+x\right)}{32}.$$
So, $$\frac{100\left(32+x\right)}{32}=125.$$
or,$$100x+3200=4000.$$
or,$$100x=800.$$
or,$$x=8.$$
A is correct choice.
A vessel of capacity V liters can be filled by two taps A and B independently in $$\frac{1}{4}$$ hr and $$\frac{1}{6}$$ hr respectively. A tap C empties the full tank at the rate of 7 liters per min. If all the 3 taps are opened simultaneously, the full vessel is emptied in 120 min. Then V =
Solution
A can fill tank in(1/4)hr=15 min.
and B can fill tank in (1/6)hr=10 min.
C can empty the 7 litter in 1 min.
So, in 1 min A can fill V/15 part of tank.
in 1 min B can fill V/10 part of tank.
1 min A,B and C can empty (7-V/15-V/10)
=(210-5V)/30.
=(42-V)/6.
=(7-V/6).
in 120 min it will empty (840-20V).
So, (840-20V)=V.
or,21V=840.
or,V=40.
So,D is correct choice.
A pipe can fill an empty cistern with water in 5 hours. Due to leakage in its bottom, it takes 6 hours to fill the cistern. When the cistern is full, the time (in hours) in which it is emptied due to leakage is
Solution
Let us call the Pipe A.
Pipe A fills the cistern in 5hrs
so in one hr Pipe A fills 1/5th of the cistern.
However with the leak it takes 6hrs for Pipe A to fill the cistern.
Hence in one hour the cistern will have 1/6th left in.
So the difference between not leaking and leaking is 1/5 - 1/6= 1/30th of the cistern
this is amount of leak per hour.
Hence it will take 30 hours to fully empty the cistern.
A is correct choice.
Three pipes A, B, C have flow rates of 2 liters, y liters and 3 liters per minute, (2 < y < 3) respectively. The lowest and the highest flow rates of the pipes are decreased by a constant quantity x. If the reciprocals of the flow rates of A, B, C are in arithmetic progression both before and after the change, then x =
Solution
According to 1st condition,
$$\left(\frac{\ 1}{y}-\frac{1\ }{2}\right)=\left(\frac{1\ }{3}-\frac{1\ }{y}\right).$$
or,$$\frac{\ 2}{y}=\frac{1\ }{3}+\frac{1\ }{2}=\frac{5\ }{6}.$$
or,$$y=\frac{12\ }{5}.$$
According to second condition,
$$\frac{1\ }{y}-\frac{1\ }{2-x}=\frac{1\ }{3-x}-\frac{1\ }{y}.$$
or,$$\frac{2\ }{y}=\frac{1\ }{3-x}+\frac{1\ }{2-x}.$$
or,$$\frac{5}{6}=\frac{3-x+2-x}{\left(3-x\right)\left(2-x\right)}=\frac{\left(5-2x\right)}{\left(x^2-5x+6\right)}.$$
or,$$\left(5x^2-25x+30\right)=30-12x.$$
or,$$5x^2=13x.$$
or,$$x=2.6.$$
A is correct choice.
A swimming pool is fitted with 3 pipes A, B, C to fill the pool. A and B together can fill the pool in half the time that is required for C to fill the pool. B takes 20 hours more than the time required for A and 14 hours more than the time required for C to fill the pool. Then the time (in hours) required for all the 3 pipes together to fill the pool is
Solution
Let say, A,B,C require a,b and c hour to fill the pool.
According to question,
b=20+a and b=c+14.
So,$$\ 20+a=c+14.$$
or,$$c-a=6.$$..........................(1)
Again, A and B together can fill the pool in half the time that is required for C to fill the pool.
So,$$\ \frac{ab\ }{a+b}=\frac{c\ }{2}.$$
Putting value from (1) and b=20+a :
$$\ \frac{a\left(20+a\right)\ }{a+20+a}=\frac{a+6\ }{2}.$$
or,$$\ \frac{a\left(20+a\ \right)}{a+10}=\frac{a+6\ }{1}.$$
or,$$\ a^2+20a=\left(a+6\right)\left(a+10\right).$$
or,$$\ a^2+20a=a^2+6a+10a+60.$$
or,$$20a=16a+60.$$
or,$$4a=60.$$
or,$$a=15.$$
So,b=35 and c=21.
So,together they will do in 1 hour$$\left(\ \frac{\ 1}{15}+\frac{1\ }{35}+\frac{\ 1}{21}\right)$$ part
or $$\frac{\ 1}{15}+\frac{1\ }{35}+\frac{\ 1}{21}=\frac{\ 7+3+5}{105}=\frac{15\ }{105}=\frac{1}{7}$$part.
So,they together will complete in 7 hour.
Mohan is thrice as efficient as Srinu and completes a work in 40 hours less than the time taken by Srinu. If both of them work together, the time (in hours) required to complete that work is
Solution
Let,Srinu takes x hour to complete work.
then Mohan takes x/3 hour to complete the work.
According to question,
$$x\ -40=\frac{x\ }{3}.$$
or,$$x\ -\frac{x\ }{3}=40.$$
or,$$\frac{2x\ }{3}=40.$$
or,$$x\ =60.$$
Then,they will do in 1 hour $$\ \frac{\ 1}{20}+\frac{1\ }{60}=\frac{3+1\ }{60}=\frac{4\ }{60}=\frac{1`\ }{15}$$ part.
So, they together will complete in 15 hour.
C is correct choice.
Two children A and B are playing a game. A can draw a picture in,30 minutes and B can erase it in 40 minutes. If A starts drawing, and if the drawing sheet is passed on to these two alternately for every one minute, then the time (in minutes) required to complete a picture for the first time is
Solution
A can do full picture in 30 min.
B can erase it in 40 min.
A's rate =$$\frac{1\ }{30}$$ per min.
B's rate=$$\frac{1\ }{40}$$ per min.
So, 1st min A will do=$$\frac{1\ }{30}$$.
2nd min work done after B erase=$$\frac{1\ }{30}-\frac{1\ }{40}=\frac{40-30\ }{120}=\frac{10\ }{120}.$$
3rd min,work done=$$\frac{10\ }{120}+\frac{1\ }{30}\frac{10+4}{120}=\frac{14\ }{120}.$$
4th min=$$\frac{14\ }{120}-\frac{1}{40}=\frac{14-3}{120}=\frac{11}{120}.$$
5th min=$$\frac{11}{120}+\frac{1}{30}=\frac{11+4}{120}=\frac{15}{120}.$$
6thmin=$$\frac{15}{120}-\frac{1}{40}=\frac{15-3}{120}=\frac{12}{120}.$$
So, every even number minutes it gain by 1 unit of work.
So, 2,4,6,......,232nd min it will complete ==> 10,11,12,.......,116th part of work.
But in 233rd min it will complete$$\frac{116\ }{120}+\frac{1\ }{30}=\frac{\ 116+4}{120}=\frac{120\ }{120}=1.$$
So, full job will done in 233 min.
D is correct choice.
18 men and 12 women can complete a work in 18 days. A women takes twice as much time as a man to complete that work. Then the number of days required for 8 men to complete the same work is
Solution
18m+12w=>18days.
2w=1m.
18m+6m=>18days.
So,24 man can do the task in 18 days.
So,8men can do it in=(24×18)/8=54 days.
B is correct choice.
A boy, a man and a woman can do a work independently in 72, 12 and 48 days respectively. The number of women required to assist 6 boys and a man to complete that work in 2 days is
Solution
A boy can do in 1 day (1/72) part of work.
A man can do in 1 day (1/12) part of work.
A women can do in 1 day (1/48)part of work.
6 boys can do in 1 day (6/72)=1/12 part of work.
let say ,x number of women needed.
So,6 boys,1 man and x women can do work in 1 day=(1/12+1/12+x/48) unit.
so, in 2 days they will do=(1/6+1/6+x/24).
So,(1/3+x/24)=1.
or,x/24=1-1/3=2/3.
or, x=16.
B is correct choice.
64 men working 8 hours a day plan to complete a piece of work in 9 days. After 5 days, they were able to complete only 40% of the work. The number of hours they should work per day so as to complete the remaining work in 4 more days is
Solution
We know ,unit of work=M×D×H
M=number of men
H=number of hour
D=number of days.
if 64 men complete 40% of work in 5 days,
then work done=64×5×8=2560 unit.
so,40% work=2560 units.
so,60% work=2560(60/40)=3840 units.
let say, to complete the work in 4 days they need to work for x hour per day .
So,64×4×x=3840
or, x=15.
D is correct choice.
Two friends A and B working together can complete a piece of work in 16 days. A alone can do the same work in 32 days. If A and B work on alternate days, starting with B, the time (days) in which the work can be completed is
Solution
A and B together can do the work in 16 days.
So, in 1 day they together can do (1/16) part of the work.
A can do (1/32) part of work in 1 day.
So,B can do $$(1/16-1/32)=1/32$$ part of work in 1 day.
So,if B do work in alternative days then they together have to work for 32 days i.e.
$$32×(1/32)=1$$.
D is correct choice.
The LCM of 96,144 and N is 576. If their HCF is 48, then a possible value of N is
Solution
Factors of the number are
$$576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^6 \times 3^2$$
$$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^5 \times 3$$
$$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$$
LCM of 96,144 and n is 576.
i.e.
$$n = 2^6 \times k = 64k$$
....(1)
HCF is 48.
$$HCF(96, 144, 64k) = 2^4 \times 3$$
$$48 = 2^4 \times 3$$
This only happen if k=3
So, substituting k=3 in (1),
$$n=64×3=192.$$
D is correct choice.
The number of zeros at the end of the product $$1003 \times 1001 \times 999 \times...\times123$$ is
Solution
To produce a 0 ,we need at least one 5 and at least one 2 .
But in the given question, all the numbers are odd number ,so there is no 2 exist in the product.
So, the above product can not produce any number of 0's.
C is correct choice.
If A = $$2^{352} 5^{411} 3^{152}$$ ; B = $$2^{352} 5^{410} 3^{153}$$ ; C = $$2^{350} 5^{412} 3^{149}$$, and D = $$2^{353} 5^{409} 3^{150}$$ then the descending order of A, B, C, D is
Solution
Let us say,$$k=2^{350}× 5^{409}×3^{149}$$
So,$$A=4×25×27k=2700k.$$
$$B=4×5×81k=1620k.$$
$$C=4×125×1k=500k.$$
$$D=8×1×3k=24k.$$
So,$$A>B>C>D$$
A is correct choice.
The smallest 5 digit number which when divided by 7,11 and 21 leaves the remainder 3 in each case is
Solution
We know that -
Dividend=divisor*quotient+remainder
For our question dividend must be of 5 digit.Let dividend be x.
Divisor=LCM(7,11,21)=231.
Let quotient be y.
Remainder is 3.
Substitute above values in equation-
x=231*y+3
Now we need to substitute different value of y(y=1,2,3…) in order to find smallest 5 digit x.
When y=1 then x=234
.
.
.
When y=43 then x=9936
When y=44 then x=10167
When y=45 then x=10398
So least 5 digit number(10167) is obtained at y=44.
The 5 digit number when divided by 7,11,21 which gives remainder 3 is 10167.
So,Option B is correct choice.
If 937x8y7 is exactly divisible by 11, then the maximum value of x + y is:
Solution
As per divisibility rule of 11,
$$9+7+8+7=3+x+y$$
or,$$31=3+x+y$$
or,$$x+y=28$$
but ,x and y are single digit number they can take a maximum value of 9 .
if both the digits are 9 then it can't be divisible.
So,either x or y is equal to 9 or 8.
So, the maximum value of x+y is 17.
D is correct choice.
If the number obtained after subtracting x from 2035 leaves the same remainder 5 when it is divided by 9,10 and 15, then the smallest possible x is
Solution
If we substract 50 from 2035 it will become
1085.
Now when 1085 is divided by 9,10 and 15 they will give a remainder of 5 each.
A is correct choice.
$$ 1.\overline{27} + 0.\overline{94} $$
Solution
Let say,
$$x=\overline{.27}\ and\ y=\overline{.94\ }\ .$$
So,we can rewrite it as,
$$x=.27\overline{27\ }\ and\ \ y=.94\overline{94\ }\ .$$
or,$$100x=27.\overline{27\ }\ and\ \ 100y=94.\overline{94\ }\ .$$
or,$$100x=27+.\overline{27\ }\ and\ \ 100y=94+.\overline{94\ }\ .$$
or,$$100x=27+x\ and\ \ 100y=94+y\ .$$ (as $$x=\overline{.27\ }\ and\ \ y=\overline{.94\ }\ .$$)
or,$$99x=27\ and\ 99y=94.$$
or,$$x=\frac{\ 27}{99}\ and\ \ y=\frac{94\ }{99}.$$
So,$$1.\overline{27\ }+.\overline{94\ }=1+x+y.$$
or,$$1.\overline{27\ }+.\overline{94\ }=1+\frac{27\ }{99}+\frac{94\ }{99}.$$
or,$$1.\overline{27\ }+.\overline{94\ }=\frac{220\ }{99}=2.22222222.....=2.\overline{2\ }.$$
B is correct choice.
Let A = {(a, b, c)/ $$c^2$$ = $$a^2 + b^2$$ }. If (3, 5, x), (y, 3, 7), (1, z, 5) are three elements of the set 'A' and the LCM of $$x^2, y^2, z^2$$ is $$p_1^{\alpha_1} p_2^{\alpha_2} p_3^{\alpha_3} p_4^{\alpha_4}$$ where $$p_1, p_2, p_3, p_4$$ are primes, then $$\frac{p_1 + p_2 + p_3 + p_4}{\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4}$$ =
Solution
It is given that ,$$c^2=a^2+b^2$$.
Now,if you compare set(a,b,c) with set(3,5,X) then we can say that a=3,b=5 & c=x.
So,$$x^2=3^2+5^2$$
or,x=√(9+25)=√34.
Similarly,
$$y=√(49-9)=√40$$ and
$$z=√(25-1)=√24$$
so,$$LCM(x^2,y^2,z^2)=LCM(34,40,24)$$
$$=2^3×17×5×3.$$
so,$$p1+P2+P3+p4=2+17+5+3=27.$$
and $$a1+a2+a3+a4=3+1+1+1=6$$
so,ratio would be$$=27/6=9/2.$$
A is correct choice.
If the number of numbers between 100 and 1000 that are divisible by 11 is x, then the number of total divisors of x is
Solution
Between 100 and 1000 there are 81 numbers present which are divisible by 11.
81 can be divisible by 1,3,9,27 & 81.
So, value of x is 5 .
B is correct choice.
Match the items of the following lists.
| List - A | List - B |
| a) a, b are prime numbers | i) LCM of $$a, b \leq ab$$ |
| b) a, b are composite numbers | ii) Conjugate surds |
| c) $$ 1.34 \overline{54}$$ | iii) Irrational numbers |
| d) $$(\sqrt[3]{2} + 3\sqrt{5})(\sqrt[3]{2} - 3\sqrt{5})$$ | iv) Rational numbers |
| v) Co-prime numbers |
Correct answer for a, b, c, d is
Solution
a) a,b are prime numbers then it will go with option(v).
b) if a and b are composite number the their multiplication shall be greater than their LCM.so,it will go with (i).
c) c will definitely go with (iv).
and d) with (iii) .
C is correct choice.
Let $$p_1, p_2, p_3$$ be prime numbers and $$\alpha, \beta, \gamma$$ be positive integers. If $$p_1^\alpha p_2^\beta p_3^\gamma$$ is a divisor of 34864764 lying between 100 and 200, then ($$p_1 + p_2 + p_3$$)($$\alpha + \beta + \gamma$$) =
Solution
$$34864764=2^2\times3\times11\times264127.$$
So,$$p_1,p_2,p_3\ could\ have\ values\ 2,3,11.$$ and $$\alpha\ ,\beta\ ,\gamma\ \ could\ have\ values\ of\ 2,1,1.$$
So,$$\left(p_1+p_2+p_3\right)\left(\alpha\ +\beta\ +\gamma\ \right)=\left(2+3+11\right)\left(2+1+1\right)=16\times4=64.$$
A is correct choice.
If x = $$\frac{1}{2 + \frac{1}{2 +\frac{1}{2 + .....\infty} }}$$ then x =
Solution
From the given question we can say that,
$$x=\frac{1\ }{2+x}\left(as\ the\ series\ reapeted\ from\ the\ same\ position\right).$$
or,$$x^2+2x=1.$$
or,$$x^2+2x-1=0.$$
or,$$x^2+2x+1-2=0.$$
or,$$\left(x+1\right)^2=2.$$
or,$$x+1=\sqrt{2\ }.$$
or,$$x=\left(\sqrt{2\ }-1\right).$$
So, C is correct choice.
The value of $$\frac{1}{1^2.3^2} + \frac{2}{3^2.5^2} + \frac{3}{5^2.7^2} + \frac{4}{7^2.9^2} + ...... + \frac{15}{29^2.31^2}$$ is
Solution
from given series we can say that,
$$\ T_n=\frac{n\ }{\left(2n-1\right)^2\left(2n+1\right)^2}=\frac{1\ }{8}\left(\ \frac{\ 1}{\left(2n-1\right)^2}-\frac{1\ }{\left(2n+1\right)^2}\right),\ where\ n=1,2,3,...,15.$$
So,$$S_n=$$
$$\ T_1+T_2+....+T_{15}$$
$$=\frac{1\ }{8}\left(\ \frac{\ 1}{1^2}-\frac{1\ }{3^2}\right)+\frac{1\ }{8}\left(\frac{1\ }{3^2}-\frac{1\ }{5^2}\right)+.....+\frac{1\ }{8}\left(\frac{1\ }{29^2}-\frac{1\ }{31^2}\right).$$
$$=\frac{1\ }{8}\left(1-\frac{1\ }{31^2}\right).$$
$$=\frac{1\ }{8}\left(\ \frac{\ 961-1}{961}\right).$$
$$=\frac{1\ }{8}\left(\ \frac{\ 960}{961}\right).$$
$$=\frac{\ 120}{961}.$$
B is correct choice.
If $$x = \frac{3 + \sqrt{6}}{5\sqrt{3} - 2\sqrt{12} - \sqrt{32} + \sqrt{50}}$$, then $$\frac{x^4 - 1}{x^4 + 1}$$ =
Solution
$$5\sqrt{3\ }-2\sqrt{12\ }-\sqrt{32\ }+\sqrt{50\ }=5\sqrt{3\ }-4\sqrt{3\ }-4\sqrt{2\ }+5\sqrt{2\ }=\sqrt{3\ }+\sqrt{2\ }.$$
$$3+\sqrt{6\ }=\sqrt{3\ }\left(\sqrt{3\ }+\sqrt{2\ }\right).$$
So,$$x=\sqrt{3\ }.$$
So,$$\ \frac{\ x^4-1}{x^4+1}=\frac{9-1\ }{9+1}=\frac{8\ }{10}=\frac{4\ }{5}.$$
C is correct choice.
Each mango costs Rs.5 and each orange costs Rs.7. If a person spends Rs.38 on these two varieties of fruits, then the sum of the number of mangos and oranges purchased by that person is
Solution
Person buys 2 mangoes and 4 oranges ,Total worth of$$=2\times5+4\times7=38.$$
A is correct choice.
If $$x = \frac{1}{\sqrt{13} - 3}, y = \frac{1}{\sqrt{7} - \sqrt{3}}, z = \frac{1}{\sqrt{2}(\sqrt{3} - 1)}$$, then
Solution
$$x=\ \frac{\ 1}{\sqrt{13\ }-3}=\ \frac{\ \sqrt{13\ }+3}{13-9}=\ \frac{\ \sqrt{13\ }+3}{4}=1.65.$$
$$y=\ \frac{\ 1}{\sqrt{7\ }-\sqrt{\ 3}}=\ \frac{\ \sqrt{7\ }+\sqrt{3\ }}{7-3}=1.09.$$
$$z=\ \frac{\ 1}{\sqrt{6\ }-\sqrt{2\ }}=\ \frac{\ \sqrt{6\ }+\sqrt{2\ }}{4}=0.96.$$
So, $$x>y>z.$$
C is correct choice.
The smallest of the differences between the perfect sqares lying on either side of the least positive integer that is divisible by 3, 4, 5, 6, 8 is
Solution
LCM(3,4,5,6,8)=120.
So, Smallest number that is divisible by 3,4,5,6,8 is 120.
Now, Smallest square numbers ,which are present on both side of 121, that can produce smallest difference are 100$$\left(10^2\right)$$ and 121$$\left(11^2\right)$$.
So, Smallest difference is (121-100)=21.
D is correct choice.
If $$x = \sqrt{2} + \sqrt[3]{5}$$ and $$y$$ is such that $$xy$$ is rational, then a value of $$y$$ is
Solution
If x is rational number then 1/x is also a rational number.
A is correct choice.
If the mean proportional of $$b, c$$ and the $$4^{th}$$ proportional of $$a, b, c$$ are both equal to 8, then $$abc$$ =
Solution
mean proportion of b,c is$$\sqrt{bc\ }=8.$$
or,$$bc=64.$$
And, $$a:b=c:8.$$
then,$$a=\frac{\ bc}{8}=\frac{64\ }{8}=8.$$
So,$$abc=8\times64=2^9.$$
A is correct choice.
The greatest number that exactly divides 513, 1134 and 1215 is
Solution
HCF of 513,1134 and 1215 is 27.
So, C is correct choice.
If $$x = 1 + \frac{1}{2^2} + \frac{1}{2^3} + ....\infty$$ and $$y = x + \frac{1}{2} + \frac{x}{9} + \frac{1}{18} + \frac{x}{81} + \frac{1}{162} + ....\infty$$, then
Solution
for an infinite series , SUM$$=\ \frac{\ a}{\left(1-r\right)}.$$
So, $$y=\ \frac{\ x}{\left(1-\ \frac{\ 1}{9}\right)}+\ \frac{\ \ \frac{\ 1}{2}}{\left(1-\ \frac{\ 1}{9}\right)}.$$
or,$$y=\ \frac{\ x}{\left(\frac{\ 8}{9}\right)}+\ \frac{\ \ \frac{\ 1}{2}}{\left(\frac{\ 8}{9}\right)}.$$
or,$$y=\ \frac{9\ x}{8}+\frac{\ \ 9}{16}.$$
Again,
$$x=\ \frac{\ \ \frac{\ 1}{2^2}}{\left(1-\ \frac{\ 1}{2}\right)}+1.$$
or,$$x=\frac{\ 3}{2}.$$
So,$$y=\frac{\ 27}{16}+\frac{\ 9}{16}=\frac{\ 36}{16}=\frac{\ 9}{4}=x^2.$$
C is correct choice.
The number of ordered pairs (x,y) of positive integers satisfying the inequality 5x + 3y $$\leq$$ 15 is
Solution
possible values are (1,1),(1,2),(1,3) and (2,1).
So, A is correct choice.
In a class, the number of boys who can swim is one more than the number of girls who can swim. The number of girls who cannot swim is one more than the number of boys that cannot swim. The difference between number of boys who can swim and number of girls who cannot swim is two. Then which of the following is true?
Solution
Let say,
number of boys who can swim is a and who cant is b.
And number of girls who can swim is c and who cant is d.
The number of boys who can swim is one more than the number of girls who can swim a = c+1
The number of girls who cannot swim is one more than the number of boys that cannot swim => d = b+1
The difference between number of boys who can swim and number of girls who cannot swim is two => a - d = 2
=> Substituting a and d values in the above equation we getc+1 - (b+1) = 2
c+1-b-1 = 2
=> c-b = 2
A book costs Rs.35 and a pen costs Rs.20. A person has purchased less number of books than the number of pens by spending maximum amount out of Rs. 350, that he has. The amount left unspent (in rupees) is
Solution
Let say,he bought x number of pens and y number of books.
each pen cost 20 rs.and each book cost 35 rs.
So,it will cost a total of $$(20x+35y)$$rs.
So, According to the question,
$$350\ge(20x+35y). $$
by using trial and error method,
if we put x=10 and y=4 then,
$$20x+35y=200+140=340$$
which is less than 350.
But if we put x=14 and y=2 then,
$$20x+35y=280+70=350.$$
So, we can utilize the 350 Rs. full only when he purchase 14 number of pen and 2 number of books.
So, no rupees will left with him.
A is correct choice.
Which one of the following point does not lie in the region bounded by x + y < 5, 2x - 4y >1 and x < y is
Solution
if we put $$\left(-\ \frac{\ 1}{10},+\ \frac{\ 1}{10}\right)$$ in $$2x\ -\ 4y>1$$,it does not satisfy the equation.
So, C is correct choice.
If A gives Rs.30 to B, then B will have twice the money left with A. If B gives Rs.10 to A, then A will have thrice as much money as is left with B. Then the amount A has initially (in rupees) is
Solution
Let say, initially A had a Rs and B had b Rs.
from condition 1 :
$$B+30=2\times\ \left(A-30\right).$$
or,$$2A-B=90.$$
from condition 2:
$$A+10=3\times\ \left(B-10\right).$$
or,$$3B-A=40.$$
By solving these 2 equations we get,
$$A=62$$ and $$B=34$$.
So, C is correct choice.
The number of leap years in between the years 2018 and 2126 is
Solution
Here first term ,a$$=2020$$ and common difference, d$$=4$$.
and last term$$=2124.$$
So,$$a+(n-1)d=l$$
or,$$2020+(n-1)4=2124$$
or,$$(n-1)4=104$$
or,$$n=27.$$
So,B is correct choice.
If $$15^{th}$$ January was Monday in the year 1952, then the last day of that year was
Solution
If we consider a 7 day cycle then next day should be a Monday.
Here 15th of January is Monday.
But 1952 is a leap year .
So there would be one extra day
Which should be next day of the week.
So Tuesday is the answer.
C is correct choice.
Which one of the following is not true?
Solution
Formula to calculate day :
(year code+Month code+Century code+Date Number-Leap year code) mod 7.
Year code:
$$(YY+(YY\div4)) mod 7$$
Here,for 1600th year,
$$(00+(00/4))mod7$$=00.
year code is 0.
Month code:
from January to December the month codes are as : 033614625035(we know it).
here month code is 5.
Century Code:
Century code for 1600 years is 6.
So, required date would be at=(0+5+6+31-0)mod 7=0(as 1600 is a leap year and month is not in January or February so we will subtract 0).
So, it will be Sunday not Saturday.
D is correct choice.
In 12 hours, how many times the hours and minutes hands of a clock will coincide with- each other?
Solution
Hour hand covers 360° in 12 hour or 720 minutes.
Also, Minute hand covers 360° in 60 minutes.
So, hour hand covers (1/2)° in 1 minute.
and minute hand covers 6° in 1 minute.
Let say, time in the clock when they coincide : hour hand moved H hour and minute hand moved M minutes.
So,
Angle between hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past HH 'o clock
$$=(30H-6M)+(M/2)°$$
$$=(30H-(11M/2))°$$
When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past HH 'o clock
$$=((6M−30H)−M/2)°$$
$$=((11M/2)-30H)°$$
So, taking mod:
$$|60H-11M|=2\theta$$..........................(1)
So, hour hand and minute hand will coincide 11 times :
between 1 and 2 , 2 and 3 , 3 and 4, 4 and 5 , 5 and 6, 6 and 7 , 7 and 8 , 8 and 9, 9 and 10, 10 and 11, 11 and 12.
1 and 2 :
between 1 and 2 hour hand will pass 1 O'clock,
So,H=1 and angle should be 0.
using equation (1):$$60\times\ 1-11M=0\ or\ M=60\div11=5\frac{5\ }{11}.$$
So,between 1 and 2 , hour hand and minute hand will coincide at $$5\frac{5\ }{11}$$ minutes past 1.
2 and 3:
$$60\times2-11M=0\ or\ M=120\div11=10\frac{10\ }{11}.$$
They will coincide $$10\frac{10\ }{11}$$ minutes past 2.
3 and 4:
$$60\times3-11M=0\ or\ M=\frac{180}{11}=16\frac{4\ }{11}.$$
They will coincide $$16\frac{4\ }{11}$$ minutes past 3.
4 and 5:
$$60\times4-11M=0\ or\ M=\frac{240}{11}=21\frac{9\ }{11}.$$
They will coincide $$21\frac{9\ }{11}$$ minutes past 4.
5 and 6:
$$60\times5-11M=0\ or\ M=\frac{300}{11}=27\frac{3\ }{11}.$$
They will coincide $$27\frac{3\ }{11}$$ minutes past 5.
6 and 7:
$$60\times6-11M=0\ or\ M=\frac{360}{11}=32\ \frac{\ 8}{11}.$$
They will coincide $$32\ \frac{\ 8}{11}$$ minutes past 6.
7 and 8:
$$60\times7-11M=0\ or\ M=\frac{420}{11}=38\frac{2\ }{11}.$$
They will coincide $$38\frac{2\ }{11}$$ minutes past 7.
8 and 9:
$$60\times8-11M=0\ or\ M=\frac{480}{11}=43\frac{7\ }{11}.$$
They will coincide $$43\frac{7\ }{11}$$ minutes past 8.
9 and 10:
$$9\times60-11M=0\ or\ M=\frac{540}{11}=49\frac{1\ }{11}.$$
They will coincide $$49\frac{1\ }{11}$$ minutes past 9.
10 and 11:
$$60\times10-11M=0\ or\ M=\frac{600}{11}=54\frac{6\ }{11}.$$
They will coincide $$54\frac{6\ }{11}$$ minutes past 10.
11 and 12:
$$60\times11-11M=0\ or\ M=\frac{660}{11}=60.$$
So, last time they will coincide 60 minutes past 11O'clock which means at 12 noon.
So, Total 11 times they will coincide.
B is correct choice.
At what time between 3 O'clock and 4 O'clock will the hands of a clock lie on a straight line but do not coincide?
Solution
At the time 3:00 O'clock hour hand is 15 minutes away from centre.
But , minutes hand gain 5 minutes in every 60 minutes.So,to be in staright line it should cover $$(15-5)=10 minutes$$ away from 4 O'clock hand.
and in 60 seconds it will gain 55 seconds.
So, They will be in a straight line at
$$4 O'clock- 10 minutes-55seconds$$
at 3:49:05.
A is correct choice.
A clock is set right at 8AM on 1.1.2018. The clock gains 15min in 24hours. The time on the clock when the actual time is 10 AM on 4.1.2018, is (nearly)
Solution
It gain 15 minutes in 24 hour or 1440 minutes.
or,in 60 minutes it gain 0.625 minutes.
so,From 8AM of 1.1.2018 to 8AM 4.1.2018, the 3 days have been covered .
So,$$(15×3)=45minutes$$ shall count.
Now from 8AM to 10AM ,it is 2 hour.
So,the clock will gain $$(0.625×2)=1.25$$ minutes.
So,it will show 10:46 AM in the clock.
C is correct choice.
A well of 4m diameter and 35m deep is dugout and the excavated soil is transported in a rectangular parallelopiped shaped truck with dimensions $$5m\times2m\times0.5m$$. To avoid over during transportation only 80% of its capacity is filled. If the loose soil occupies 20% more space while filling into the truck, then the number of trips required to transport the soil completely away from the place of digging is
Solution
Volume of well$$=π2^2×35=440m^3.$$
Volume of truck$$=5×2×0.5=5m^3.$$
but the truck is only able to fill its 80% of capacity to fill the space.
So, effective amount was$$=0.8×5=4m^3.$$
So,20% of wasted amount is
$$=0.2×440=88m^3.$$
So, number of trips required are
$$=(440/4)+(88/4)=110+22=132.$$
D is correct choice.
If the volumes of a sphere and a cube are in the ratio $$9\pi$$ : 2, then the ratio of the radius of the sphere to the edge of the cube is
Solution
Let say, Volume of sphere$$=9πk.$$ and Volume of cube$$=2k.$$
So,$$(4/3)πr^3=9πk$$
or,$$r=3k/2^(2/3).$$
and edge of cube$$=k2^(1/3).$$
So, required ratio would be$$=3:2.$$
B is correct choice.
The dimensions of a rectangular plot is $$40m\times20m$$. A path is formed across the plot along the length and breadth with a uniform width of 2 meters. If the cot of forming, the path is Rs. 500 per sq.meter, the cost of laying that path (in Rs.) is
Solution
According to the question,
length of the path would be 40m and width of the path would be 2m by the length side(area of that path is$$=40×2=80m^2$$).
And ,length of the path would be 20m and width of the path would be 2m by the breadth side(area$$=20×2=40m^2$$).
If it cost 500 Rs. to laying the path, total cost of laying the path would be
$$=80×500+40×500=60000.$$
But,in corner a reapetated cost is taken during the path ,whose length and breadth would be of 2m respectively.
So,area of that portion is$$=2×2=4m^2.$$
So,cost of that path$$=4×500=2000Rs.$$
So,Total cost$$=60000-2000=58000.$$
A is correct choice.
A person packs sweets boxes of dimensions $$15cm\times15cm\times15cm$$ in a basket of size $$120cm\times120cm\times120cm$$. If he now wants to carry the maximum number of cubical boxes having the maximum integer dimensions but less than the dimensions of the earlier boxes without leaving any space unused, the number of such boxes he can carry is
Solution
He can carry maximum number of boxes ,only if he put one box over another .
So, maximum number boxes he can carry is
$$=15×15×15=3375.$$
C is correct choice.
From a circle of radius 12cm centered at O, a sector OAB of are length $$8\pi$$ cm is cut and from it a cone is formed by joining OA and OB. If the volume of the cone is V cubic cm and its lateral surface area is S square cm, then V:S =
Solution
Length of the arc is$$=8π.$$
So, perimeter of the base of cone is$$2πr$$.
So,$$2πr=8π.$$
or,$$r=4.$$
radius of the circle is 12 cm.
So,slant height of the cone is 12 cm.
So, height of the cone$$=√(12^2-4^2)$$
$$=√128=8√2.$$
So,V$$=(1/3)π4^2×8√2.$$
and S$$=π×4×√(h^2+r^2)=4π×12.$$
So,V:S$$=8√2:9.$$
C is correct choice.
A solid sphere of radius $$r$$ is melted and with that material a solid cone and twenty two identical solid cubes were made. If the height of cone and edge of each cube are each equal 9 to half the radius of the sphere, then the ratio of the radius of the cone to its height is
Solution
Volume of the sphere is
$$=(4/3)πr^3.$$
height of cone r/2.
let say radius of cone is a.
So, volume of cone is
$$=(1/6)πa^2r.$$
side of cube is $$=r/2.$$
volume of cube$$=r^3/8.$$
So,$$(4/3)πr^3=(1/3)πa^2r+r^3/8.$$
So,$$a=(√11/2)r$$.
So, height: radius$$=√11:1$$.
B is correct choice.
What is the area (in sq. cms) of the shaded portion of the following diagram in which each side of the triangle is 14 cms and its vertices being the centres of three mutually touching circles?
Solution
Radius of circle is$$=14/2=7cm.$$
From the picture we can say that area of the shaded region is$$=(1/3)×3×area$$.
So, required area$$=π×7^2=154$$.
D is correct choice.
Let $$\triangle$$ be the area of the circumcircle of a right angled triangle ABC with $$\angle B$$ = 90°. Let $$\triangle_1 and \triangle_2$$ be areas of the two circle with diameters BC and BA respectively. Then
Solution
Let say ,a be the length of the side conjugate to the right angle .
so,length of the diameter of the circle is
$$=√2a$$.
so,radius$$=(√2/2)a$$.
so,$$∆=πa^2/2$$.
Now,∆1=∆2=
$$π(a/2)^2=πa^2/4.$$
So,∆1+∆2=$$πa^2/2.$$
So,∆=∆1+∆2.
C is correct choice.
From each of the corners of a rectangular sheet of dimensions $$36cm\times24cm$$, a small square of dimensions $$4cm\times4cm$$ is removed. If the edges on the four sides are folded and a box is formed then the volume of the box so formed (in cm3)
Solution
If 4 squares of 4×4 is removed from each corner then length of the rectangle become
$$36-4-4=28cm$$ , breadth of the rectangle become$$=24-4-4=16cm$$ and height of the box is 4 cm.
So,volume of box is$$=28×16×4=1792.$$
A is correct choice.
In a square of side 8 cm, 5 identical circles are placed as shown in the figure. Then the radius (in cms) of each of the circle is
Solution
So,length of the diagonal of the square=$$\sqrt{2}r+\sqrt{2}r+\left(4\times r\right)=\left(4r+2\sqrt{2}r\right)$$cm.
But ,diagonal of Square=$$\sqrt{a^2+a^2\ }=\sqrt{2}a$$=$$8\sqrt{2}.$$(square of side 8 cm)
So,
$$8\sqrt{2}=4r+2\sqrt{2}r.$$
or,$$r=\frac{8\sqrt{2}}{4+2\sqrt{2}}.$$
or,$$r=\frac{8\sqrt{2}\left(4-2\sqrt{2}\right)}{16-8}=\sqrt{2}\left(2\left(2-\sqrt{2}\right)\right)=2\left(2\sqrt{2}-2\right)=4\left(\sqrt{2}-1\right).$$
C is correct choice.
Two identical circles intersect such that their centers and their points of intersection, form a square of side 4cm. Then the area (in sq.cms) of the portion that is common to the two circles is
Solution
Side of the square is 4 cm.
So,radii of the two circles are also 4 cms.
So, area of the one of the arc is
$$(π×4^2/4)=4π.$$
Area of the square is$$=4×4=16cm^2.$$
clearly the common sector have the half of square common to it.
So,area of one arc is
$$4π-16/2=4π-8 cm^2$$
So,Total area of the arc is$$8π-16$$
$$=8(π-2).$$
A is correct choice.
The number of Positive integers x such that 2018, 1998 and x are the sides of a triangle is
Solution
As 2018,1998 and x are the side of triangle.
Then sum of two sides should be greater than the third side.
So,$$2018+1998>x$$
or,$$4016>x.$$
revesely,$$1998+x>2018.$$
So, minimum value of X should be 21 .
And maximum value of X should be 4016.
So,x can take$$(4016-21)=3995$$ values.
D is correct choice.
An equilateral triangular plate is completely cut into n number of identical small equilateral triangular plates. Which one of the following can be a possible value for n
Solution
Minimum no. of identical small equilateral triangles that can be cut is 4. (1 at the top, 3 at the bottom). These 4 can be cut into 4 small triangles each.
Hence, the next value is 4*4 = 16 Hence, the number of identical small triangles that can be cut is of the form $$4^n. 256=4^4$$
C is correct choice.
The maximum of the distance between any two points of a unit cube is, in proper units
Solution
Largest distanc in a cube is the diagonal of the cube.
if we consider a square of unit side then the diagonal of that square is
$$=√(1^2+1^2)=√2.$$
so,value of largest diagonal is
$$=√(1^2+(√2)^2)=√3.$$
D is correct choice.
A train started at 9AM from a station A with a speed of 72 km/hr. Another train after tvvo hours started from the station B towards A with a speed of 90 km/ph. The two trains are expected to cross each other at 1.30 PM. At 12 noon because of the signals both the trains reduced their speeds by the same quantity and they crossed each other at 4.30 PM. The speed of the train, after 12 noon, that started from the station A, is
Solution
The train which started from station A traveled 144 km in first 2 hours.
The train from station A and station B had speed of 72 km/h and 90 km/h respectively.
And they expected to be meet in 1:30 PM.
But from 11:00 AM they started to travel towards eachother.It means that they expected to meet eachother by 2.5 hours from 11:00AM.
Their relative velocity is=(72+90)=162 km/h.
So, in 2.5 hours they should travel=2.5×162=405 km. So they were 405 km apart from eachother at 11:00 AM.
Let say, due to signal,they reduced their speed by x km/h.
So,train from station A and station B will have (72-x) km/h and (90-x) km/h as their respective speed.
But,they shifted their speed to this reduced speed at 12 noon, which means that they travel for more 1 hour from 11:00AM at their previous speeds.
So,in that 1 hour they travel for more 72 km and 90 km further towards eachother.
Which in turn reduced their total journey of 405 km by (72+90)=162 km by this 1 hour.
So,they traveled only (405-162)=243 km in (12 noon-4:30 PM)= 4.5 hours.
But their new relative velocity is (162-2x) km/h.
So, they will take (243/(162-2x)) hours to cross eachother.
So,
$$(243/(162-2x))=4.5$$
or,$$54=162-2x$$
or,$$2x=108$$
or,$$x=54$$.
So, the train ,which left from station A,have reduced its speed to (72-54)=18 km/h.
C is correct choice.
A left his house for school t min late from normal time and travelled with $$\frac{4}{3}$$ of his usual speed and reached the school in t minutes early. Next day he left home, t minutes early than the previous day and travelled with $$\frac{8}{11}$$ of his usual speed. He reached his school in
Two points A and B lie along a line at a distance of 120km apart. P and Q start at the same time from A and B with speeds 40km/h and 60km/h respectively. They travel towards each other and after their meeting for the first time, they then go in reverse directions and also interchange their speeds. These speeds are continued in their further journey. After reaching their respective starting points, they reverse their directions of travel and proceed towards each other again. The time taken by them to meet each other 2nd time is
Solution
First they go in opposite direction .
So, relative velocity of them is (40+60) km/h or 100 km/h.
So, time taken by them to meet for first time is =(120/100) h= 1.2 hours.
So,in their first meeting P travels =(1.2×40)=48 km and Q travels=(1.2×60)=72 km.
But,after first meeting P and Q reversed their speeds into 60 km/h and 40 km/h respectively.
Now,P will take (48/60)=0.8 hours to reach point A and Q will take (72/40)=1.8 hours to reach point B.
But in the mean time, P will move 60 km further from point A towards point B in next 1 hour when Q was traveling towards B.
So far ,after first meeting they traveled for 1.8 hours to get a head on move towards eachother.
So,for their second meeting they had to travel for only (120-60)=60 km.(as P already moved 60 km from point A)
So,100 km/h would be their relative velocity.
So,they will take (60/100)=0.6 hours to meet for the second time.
So,Total time taken by them to meet for the second time is=(1.2+1.8+0.6)=3.6 hours.
B is correct choice.
A person rows a distance of 12 kms in down stream and returns to the starting point. The difference between the times taken to travel in down stream and that of upstream is 6 hours. If he doubles his speed throughout the above trip, then the difference between the times taken to cover in down stream and upstream is 1 hour. Then the speed of the current in km per hour is
Solution
Let say, speed of the boat is b km/h and speed of the current is c km/h.
So, in downstream speed is (b+c) km/h and in upstream speed is (b-c) km/h.
Time taken in downstream=(12/(b+c)) h.
and time taken in upstream=(12/(b-c)) h.
So, according to question:
$$(12/(b-c))-(12/(b+c))=6.$$
or,$$(b+c-b+c)/(b^2-c^2)=(1/2).$$
$$b^2-c^2=4c.$$
$$b^2=c^2+4c.$$..............(1)
if speed of speed is doubled ,then speed in downstream =(2b+c) km/h .
and speed in upstream=(2b-c) km/h.
So, time taken in downstream 12/(2b+c) h.
And time taken in upstream 12/(2b-c) h.
So,
$$12/(2b-c)-12/(2b+c)=1$$
or,$$(2b+c-2b+c)/(4b^2-c^2)=(1/12)$$
or,$$24c=4b^2-c^2$$
or,$$b^2=(c^2+24c)/4.$$........(2)
So,from (1) & (2) :
$$(c^2+24c)/4=c^2+4c.$$
or,$$c^2+24c=4c^2+16c$$
or,$$3c^2=8c$$
or,$$c=8/3$$
So,speed of current is 8/3 km/h.
A is correct choice.
Certain amount is divided into 3 parts such that those parts yields the equal interest after 1, 2 and 3 years respectively at 4% simple interest per annum. Then the ratio between the three parts respectively is
Solution
Let say,
p,q and r are the three parts.
So,Then the simple interest would be
$$(p×0.04×1)=(q×0.04×2)=(r×0.04×3)$$
$$=k(let say).$$
$$p=25k,q=12.5k and r=(25/3)k.$$
So,
$$p:q:r=25k:12.5k:(25/3)k$$
$$=k:0.5k:(k/3)$$
$$=6:3:2.$$
A is correct choice.
A man borrows Rs. 5000 at the rate of 10% compound interest per annurn• At the end of each year he pays back Rs. 1500 for the first 2 years. The am ount he should pay at the end of 3rd year to clear the loan (in Rs.) is
Solution
At the end of first year
$$=5000×1.1=5500$$Rs would be paid.
After 1500 payment,he would be left with
$$5500-1500=4000.$$
At the end of second year
$$4000×1.1=4400.$$
After 1500 payment,he would be left with
$$4400-1500=2900.$$
So,at the end of third year the amount would become$$=2900×1.1=3190.$$
B is correct choice.
The equal annual installment which clears the debt of Rs.7620 due in 3 years at $$16\frac{2}{3}$$% per annum compounded is (in Rs.)
Solution
Let say ,x is the equal installment to be paid.
So,
$$7620(1+1/6)^3= x(1+1/6)^2$$
+x(1+1/6)+x.
or,$$x=3430(approx).$$
So,A is correct choice.
A tree is growing in height at the rate of 10% per annum. The present height of the tree is 1815 cm. The height of the tree 3 years back was (approximately) in cms
Solution
Let say, 3 years back height of the tree was p.
So,$$p×(1.1)^3=1815.$$
or,$$p=1363.66$$
or approximately 1364 cms.
So, B is correct choice.
Two partners invested Rs.1,25,000 and Rs.85,000 respectively in a business. They agreed to share 60% of the profit equally and the remaining profit as interest on their respective capitals. If one partner gets Rs.300 more than the other partner, the profit on the business is (in Rs)
Solution
Let say ,total profit earned is k.
So,they will divide 0.60k equally among them ; each will get 0.30k.
And rest of the profit, i.e.0.40k will divide in the ratio of$$125000:85000=25:17.$$
So,one will get $$0.40k×(25/42)$$ and other one will get$$0.40k×(17/42)$$.
So,$$0.40k(25/42-17/42)=300.$$
or,$$(8/42)k=750.$$
or,$$k=3937.50$$
So, D is correct choice.
Let A and B enter into a partnership with capitals in the ratio 5:6 and at the end of 8 months A withdrew from the business. If they shared the profits in the ratio 5:9, the number of months B's capital remained in the business is
Solution
Let say capital invested by A and B are 5k and 6k respectively.
So,profit ratio would be
$$=5k×8:6k×12$$
$$=5:9.$$
So, B's amount remained in the business for 12 months.
D is correct choice.
A, B and C enter into a partnership by pooling the capital. A advances one-third of the capital for one third of the time. B advances half of the capital for half of the time. C advances the remaining part of the capital for whole the period. The ratio in which they divide the profit obtained in the business is
Solution
Let say,total capital is C and total time is T.
A invested$$=C/3.$$ for T/3 period of time.
B invested$$=C/2.$$ for T/2 period of time.
C invested$$=C-C/2-C/3=C/6.$$ for T period of time.
Their profit ratio should be as follows:
$$(C/3)(T/3):(C/2)(T/2):(C/6)T$$
$$=(1/9):(1/4):(1/6)$$
$$=4:9:6.$$
So, C is correct choice.
A, B, C started a business by investing Rs.6500, Rs.8400 and 10,000 respectively. As working partner A takes 5% of the profit as his salary. A withdraws his capital after 6 months, B after 5 months and C after 3 months from the begining. If the profit earned was Rs.7400, then the total amount received by A is (in Rs.)
Solution
A,B,C will take the profit in following ratio
$$=6500×6:8400×5:10000×3$$
$$=39000:42000:30000$$
$$=39:42:30$$
$$=13:14:10.$$
Total profit$$=7400.$$
5% of 7400$$=370.$$
rest of profit$$=7400-370=7030.$$
A will get$$=7030×(13/37)=2470.$$ of rest of the profit.
So,Total profit part of A$$=2470+370.$$
$$=2840.$$
So,C is correct choice.
If the difference between the simple interest and compound interest for two years on a certain sum at the rate of 10% per annum when compounded half yearly is Rs. 124.05 then the sum is (in Rs.)
Solution
Let say Amount is P.
Compound interest
$$=P(1+r/n)^nt -P $$
and
Simple interest
$$=Pnr.$$
So,CI$$=P(1+0.10/2)^4-P=0.21550625P$$.
And SI$$=P×2×0.10=0.2P$$.
So,$$0.21550625P-0.2P=124.05$$
or,$$P=8000.$$
So,A is correct choice.
If an amount doubles in 5 years at compound interest, then the number of years required to make it eight times at the same rate of compound interest is
Solution
Let say,$$Amount=P$$ and
$$rate of interest=r percent$$
So, according to the question,
$$P(1+r/100)^5=2×P$$.
$$or,(1+r/100)^5=2$$
So,$$r=14.86 percent$$
let say ,in x years it will become 8 times.
So,
$$P(1+r/100)^x=8P$$
or,$$(1+r/100)^x=2^(15/5)$$
or,$$(1+r/100)^x=(1+r/100)^15$$
So, x=15.
D is correct choice.
A person invested his money in the three schemes P, Q and R. The am_ ount invested it scheme R was 150% of the amount invested in a scheme P and 240% of the amount invest; in scheme Q. The rate of interest he gets from schemes P, Q and R are respectively ioZ 12% and 15% per annum. If the total interest accumulated is Rs.3200 for an year, investment in scheme Q is (in Rs.)
Solution
Let say he invested in scheme P,Q and R is p,q,r respectively.
So,$$r=1.5p and r=2.4q$$
so,$$1.5p=2.4q$$
or,$$p/q=24/15=8/5$$
Let say, $$p=8k and q=5k$$
So,$$r=12k.$$
So,
$$(8k×0.10+5k×0.12+12k×0.15)=3200$$
$$or,(0.8k+0.6k+1.8k)=3200.$$
or,$$3.2k=3200$$
or,$$k=1000.$$
So,amount invested in scheme Q was
$$=5×1000=5000.$$
B is correct choice.
A person invest certain amounts in two banks in such a way that the simple interest fro one bank at 10% per annum for 5 years is equal to that from another bank at 9% per annul/ for 6 years, then the ratio between the two amounts
Solution
Simple interest from 10% interest rate for 5 years is$$=a×0.10×5=0.50a.$$(a is the principal amount).
Simple interest from 9% interest rate for 6 years is$$=b×0.09×6=0.54b.$$(b is the amount invested in this case)
According to the question,
$$0.50a=0.54b$$
or,$$a/b=54/50=27/25.$$
So, C is correct choice.
Read the following graph and answer the questions below
The diagram below shows the enrolment of students in a school in different years, from 2014 to 2017.
The total number of students enrolled in the years 2014, 2016, 2017 put together is
Solution
Total number of students in 2014,2016&2017 put together is
$$=200+250+450=900.$$
C is correct choice.
The percentage increase in the enrolment of students in the year 2017 over the year 2016.
Solution
In the year 2016 & 2017 ,number of students was 250&450 respectively.
So, percentage increased
$$=(450-250)/250=80 percent.$$
A is correct choice.
The ratio of enrolment of students in the year 2015 to that in the years 2014 and 2016 together is
Solution
In the year 2015 , number of students was 400.
In the year 2014 & 2016 , number of students was 200&250 respectively.
So, required ratio is$$=400:(200+250)$$.
$$=400:450=8:9.$$
B is correct choice.
The percentage increase in the enrolment of students during the given period is
Solution
In the year 2014 number of students was 200 & In the year 2017 number of students was 450.
So, percentage increase is
$$=((450-200)/200)×100=125 percent$$
So, D is correct choice.
The average of the number of students admitted per year during the given period is
Solution
In the year 2014,2015,2016 & 2017 number of students was 200,400,250 & 450 respectively.
So, average number of students was
$$=(200+400+250+450)/4=325.$$
So,D is correct choice.
Read the following Table and answer the questions below
The following table shows the sales of cars of different models by a company during the years 2013 to 2017.
In the year 2016, what is the approximate percentage of sales of cars of type 'C' in the total
number of cars sold in that year?
Solution
Required percentage$$=(138/785)×100$$
$$=17.57 percent$$.
D is correct choice.
During the given period, the car model that has registered highest growth percent in sales is
Solution
D has the highest growth percent during the period which is $$=(164-44)/44=2.72$$.
C is correct choice.
The year in which cars of model 'D' registered highest growth percent in the sales, over the previous year is
Solution
In the year 2016 the company sale was 207 which was 102 in 2015 for D.
So in the year it attains over 100 percent growth in the year 2016.
So, option C is correct choice.
The year in which the company registered highest growth percent in the total sales of cars, over its previous year is
Solution
In the year 2014 company occurs a loss in total sales.
In the year 2015 company gained
$$=(775-734)/734=5.58 percent.$$
In the year 2016 company gained
$$(785-775)/775=1.29 percent.$$
In the year 2017 company gained
$$=(1005-785)/785=28.02 percent.$$
So, in 2017 company gained highest percentage.
So, A is correct choice.
The approximate percentage increase in the sales of Cars of models A and D put together in the year 2016 over the year 2013 is
Solution
In the year 2013 total sales of A and D was $$=88+44=132.$$
In the year 2016 total sales of A and D was $$=207+175=382.$$
percentage change during the period was$$=(382-132)/132=1.89 $$ or 189 percent.
So, C is correct choice.
Fill in the blanks with suitable positive integer in each of the questions.
Solution
Pattern is as follows:
$$(4-1)=3 , (8-5)=3.$$
$$(9-7)=2 , (6-4)=2.$$
Similarly,
$$(8-4)=4 , (6-2)=4.$$
B is correct choice.
Solution
Given series follows following pattern:
$$5^2+3^2+2^3+1^3=43.$$
$$6^2+4^2+2^3+3^3=87.$$
Similarly,
$$7^2+8^2+2^3+4^3=185.$$
So, D is correct choice.
Solution
Series follows following pattern:
$$7×6+5=47$$
$$8×6+4=52$$
So, $$7×4+9=37.$$
A is correct choice.
Solution
Series follows following pattern:
$$5^3-6^2=89$$
$$6^3-7^2=167$$
So, $$8^3-9^2=431.$$
So, B is correct choice.
For the following questions answer them individually
Digits 1 to 6 are marked one each on the faces of a die. 3 different positions of a die are 3 given below. The combination of digits on the faces opposite to each other are.
Solution
2 is adjacent to 6&3 and 5 is adjacent to 6&3 ,which indicates that 2 is opposite to 5.
Now 4 is adjacent to 3&5 ,so , 4 is definitely opposite to 6..Then 3 is opposite to 1.
C is correct choice.
All the faces of a cuboid arc painted. It is cut by planes parallel to its faces to form unit cubes. If the number of cubes having all unpainted faces is 1001, then the dimensions of , the cuboid are
Solution
Option A is correct choice.
Because ,if we cut the cube by $$13×9×15$$ dimensions,then all the faces with the color will be out side faces of the cuboid.
So, uncolored face will stay only inside the cuboid with $$(13-2)×(9-2)×(15-2)$$ or $$11×7×13$$ or 1001 number of small cuboid.
A is correct choice.
The minimum number of different colours required to paint the surfaces of a cuboid so that no two adjacent faces are painted with same colour, is
Solution
If we want to color a cuboid with minimum number of colours then opposite faces of the cuboid should colored by same color.
So,a minimum of 3 different colour needed.
D is correct choice.
In the following figure, each cell marked with a symbol is a face of a cube. If the surfaces are folded to form a cube, then the faces that are adjacent to the face labelled $$\theta$$ are
Solution
If we fold the image then $$\theta$$ will be definitely opposite to $$\rho$$ sign. It means that alpha,beta,Gama and delta would be adjacent to theta.
Option A is correct choice.
Read the following information and answer the quetions below.
The principal of a college scheduled the following week, starting from Monday to Sunday.
i) Management skills six lecture
ii) Decision making
iii) Motivation
iv) Soft skills
v) Faculty development
vi) Quality circles
It is decided to organise the lecture on Motivation immediatly after Faculty development. The lecture on Quality circles is to be scheduled on Wednesday and it should be followed by Soft skills. The lecture on Decision making should be organised on Friday. The lecture on Management skills and the lecture on Soft skills are to be scheduled with a gap of two days M which no lecture is to be scheduled on one day (Saturday is not that day) just after the lecture day on Soft skills.
On which day, no lecture is arranged in a week?
Solution
A is correct choice.
How many lectures were organised between Quality circles and Motivation?
Solution
C is correct choice.
The lecture on Management skills is to be organised on
Solution
From first clue :
Motivation comes immediately after Faculty Development.
So,
Quality Circles and Decision Making should be scheduled on Wednesday and Friday respectively.
So,
But Quality Circles should be followed by Soft Skills .
So, Soft skills can be scheduled on either Monday or Tuesday.
But there should be one day gap after Soft skills.
So,Soft Skills should scheduled on Monday.
Now, there is a gap of two days between Soft Skills and Management Skills.
So,Management Skills should scheduled on Thursday.
And FD & Motivation are scheduled on consecutive days ,therefore these two subjects should
scheduled on Saturday & Sunday i.e. one after another.
So,Final arrangement would be :
D is correct choice.
For the following questions answer them individually
The second figure in the first pair of problem figures bears a certain relationship to the first figure. Similarly, one of the figures in the answer figures bears the same relationship to the first figure, in the second pair of the problem figures. You have to locate the figure which would fit into the blank space and give it as your answer.
Solution
If we compare part A and C in problem figure,
there is a white colored shape with 4 side in the base on which a black colored base is placed with 3 side.
So, similarly, image in part B should exchange with (C)image in answer figure,
where base part should be 3 side shape with black color on which a white color shape with 4 side should place.
B is correct choice.
Select the related figure from the answer figures that fits with the blank space of the following analogy, and give it as your answer.
Solution
In the given series,
2 triangles are concatenated one over another by rotating in reverse manner. Similarly,
if we add 2 shape by rotating then the diagram in Answer figure A will appear as the picture in the given series.
So, Option C is correct choice.
Based on following sequence of numbers, letters and symbols, answer the questions
$$\alpha$$ * S A 3 $$\beta$$ 7 C 5 @ L $ P 4 $$\uparrow$$ Q M 2 L B & 4 9 D E I X Y # 6
The number of letters of English alphabet that are immediately followed by a number or immediately peceeded by a symbol is
Solution
S,A,C,L,P,Q and M are the 7 letters satisfy the given condition.
So, B is correct choice.
The number of symbols either preceeded or followed by a number is
Solution
Beta,@,upper arrow,& and # are 5 symbols are ther in the sequence which validate the given condition.
So, B is correct choice.
If all symbols are removed from the above sequence then tha entry in 9th position from right is
Solution
If all symbols are removed from the sequence then from right side two symbols will be removed between the last letter and B.
So, B will become the 9th letter from right hand side .
So, Option A is correct choice.
The difference between the number of letters of English alphabet and the numbr of numerals in the sequence is
Solution
Number of english alphabets in the series is 14.
And Number of numerals is 8.
Difference between them is 6.
B is correct choice.
The 14th element to the left of the 6th element lying to the right of 'B' is
Solution
6th element to the right of B is "I".
And 14th element to the left of I is "$".
D is correct choice.
A sequence of numbers or alphabets following a specific pattern is given. Fill in the blanks with suitable entry from the options that follows the same pattern.
10, 30, 68, 130, ..........
Solution
Given series is in the following form,
$$2^3+2,3^3+3,4^4+4,5^5+5,6^6+6$$
D is correct choice.
3, 14, 39, 84, ..............
Solution
Given series is in the pattern of $$\left(n+n^2+n^3\right).$$
So,
$$n=1\longrightarrow\ 1+1^2+1^3=3.$$
$$n=2\longrightarrow\ 2+2^2+2^3=14.$$
$$n=3\longrightarrow\ 3+3^2+3^3=39.$$
$$n=4\longrightarrow\ 4+4^2+4^3=84.$$
$$n=5\longrightarrow\ 5+5^2+5^3=155.$$
So, C is correct choice.
$$1\frac{1}{7}, 3\frac{6}{7}, 9\frac{1}{7}, 17\frac{6}{7}, ...........$$
Solution
$$1\ \frac{\ 1}{7}=\ \frac{\ 8}{7}=\ \frac{\ 2^3}{7}.$$
$$3\frac{\ 6}{7}=\ \frac{\ 27}{7}=\ \frac{\ 3^3}{7}.$$
$$9\frac{\ 1}{7}=\ \frac{\ 64}{7}=\ \frac{\ 4^3}{7}.$$
$$\ 17\frac{\ 6}{7}=\ \frac{\ 125}{7}=\ \frac{\ 5^3}{7}.$$
$$\ 30\frac{\ 6}{7}=\ \frac{\ 216}{7}=\ \frac{\ 6^3}{7}.$$
So, B is correct choice.
14, 78, 252, 620, .......
Solution
$$2^4-2=14.$$
$$3^4-3=78.$$
$$4^4-4=252.$$
$$5^4-5=620.$$
$$6^4-6=1290.$$
So, A is correct choice.
1, 4, 10, 20, .........
Solution
f(n)$$=\frac{n\left(n+1\right)\left(n+2\right)}{6}.$$
So,$$f(1)=1$$
$$f(2)=4$$
$$f(3)=10$$
$$f(4)=20$$
$$f(5)=35$$
So,B is correct choice.
ABC, BDF, CFI, DHL, .......
Solution
$$ABC\longrightarrow\ $$ has no gap between letters.
$$BDF\longrightarrow\ $$ has one letter gap.
$$CFI\longrightarrow\ $$ has two letters gap between them.
$$DHL\longrightarrow\ $$ has three letters gap between them.
So, next word in the series is EJO(each letter has 4 gaps between them).
A is correct choice.
B, C, E, G, .......
Solution
next letter of B is C.
After one letter of C is E.
So,4 letter after G is K.
So,C is correct choice.
AZ, CW, FR, JK, .......
Solution
Using conversion table,
$$AZ=1+26=27.$$
$$CW=3+23=26.$$
$$FR=6+18=24.$$
$$JK=10+11=21.$$
So,27,26,24,21 are in difference of 1,2,3 respectively.
So,next number should be 17.
$$OB=15+2=17.$$
So, B is correct choice.
BCE, CEG, EGK, GKM, ..........
Solution
A, D, I, P, .........
Solution
$$A+3=D,\ D+5=I,\ I+7=P,\ P+9=Y.$$(using english alphabets system)
A is correct choice.
Select the suitable entry from the given options that fills the blank in the following analogies.
AFHE: IKMO:: ...... :UDFA
Solution
Using English alphabet conversion table ,difference between total of each word is same in given series.
So,IPRO is correct choice.
3265 : 4376 :: 4673 : ......
Solution
3265:4376--->
$$32+11=43 and 65+11=76$$.
Similarly,
4673--> $$46+11=57 and 73+11=84$$.
So, required number is 5784.
A is correct choice.
IMPLICATE : INCRIMINATION :: EXONERATE : ........
Solution
IMPLICATE and INCRIMINATION are similar words.
EXONERATE and ACQUITAL are similar words.
So,D is correct choice.
216 : 625 :: ....... : 4096
Solution
Given series is as follows:
$$6^3:5^4::9^3:8^4$$
implies that
$$216:625::729:4096$$.
So, B is correct choice.
LIGHT : CANDLE :: ....... : .......
Solution
CANDLE is the source of LIGHT.
Similarly, BATTERY is the source of POWER.
So,D is correct choice.
In each of the questions, a question is followed by two statements I and II. Give your answer
Who is the youngest among A, B, C?
I. The difference between the ages of A and B is 3 years.
II. A is 4 years younger to C.
Solution
Difference between the age of A and B is 3.
so we can't say anything about the value of A and B.A and B can take any integer values.
From statement 2 we can say that C is elder than A.
So,from both the statements together we are not able to find the youngest value among A,B and C.
So, D is correct choice.
If numbers x, y, z are all less than 40, how many of them are positive?
I. x + y = 46
II. x + y + z = 80
Solution
From statement I we can derive that maximum value of X or Y can have 39 and minimum value of X or Y can have 7.
So,they both are positive but we can't say anything about the value of Z.
From statement II ,we can say that any two of X,Y,Z can have maximum value of 39 then other variable will achieve minimum value of 2.
So,they all are definitely positive.
So, Option B is correct choice.
What is the value of ab?
I. a + b = 9
II. | a - b | = 5
Solution
$$|a-b|=5$$ have two equations.
1. $$(a-b)=5$$
and
2. $$(b-a)=5$$
So, either $$a=7,b=2$$or$$a=2,b=7$$.
So both the statements are required to answer the question.
C is correct choice.
What is the volume of the cone?
I. The base radius of the cone is numerically equal to the perimeter of the square ABCD.
II. Length of the side of the square is 4 cm.
Solution
Volume of cone$$=(1/3)πr^2h.$$
from statement I and II radius of the cone can be found but we can't get the value of height of the cone.
So,Option D is correct choice.
What is the angle of the sector at the centre of a circle?
I. The perimeter of the sector is 16 cm
II. Arc length of that sector is 10 cm
Solution
Perimeter of a sector is
$$=2r+length of arc$$.
So both the statements require to get the value of angle.
c is correct choice.
Is the perimeter of the rectangular plot more than 65 cms?
I. Its length is less than 16 cm
II. Its breadth is more than 10 cm
Solution
Perimeter of rectangle is$$2(l+b)$$.
and we also know that length of rectangle is greater than its breadth.
if$$2(l+b)=65$$
means $$(l+b)=32.5$$
if $$l=15$$ then $$b=17.5$$
So, whatever the value length(less than 16) takes,value of breadth is greater than length,which violates the property of a rectangle.
So, statement I is derives that such rectangle can not be possible.
So, statement I is enough to prove.
But if breadth is greater than 10 ,it can take any value between 10 and 16.
So,we can't say anything about it's perimeter.
So,Option A is correct choice.
Is the slope of the straight line equal to $$\frac{3}{5}$$ ?
I. The straight line is passing through the point (3,5)
II. The straight line perpendicular to 5x - 3y + 4 = 0
Solution
To find the slope, you divide the difference of the y-coordinates of 2 points on
a line by the difference of the x-coordinates of those same 2 points.
In statement I no coordinates are given for line.
II.
$$5x-3y+4=0.$$
or,$$5x+4=3y.$$
or,$$\ \frac{\ 5}{3}x+\frac{\ 4}{3}=y.$$
The slope of the line perpendicular to this one will have a slope equal to the negative reciprocal.
The perpendicular slope is $$-\ \frac{\ 3}{5}.$$
Plug the new slope and the given point into the slope-intercept form to find the y-intercept.
$$5=-\ \frac{\ 3}{5}\times3+b$$
or,$$\ \frac{\ 34}{5}=b$$.
$$y=-\ \frac{\ 3}{5}x+\frac{\ 34}{5}.$$
So, Option B is correct choice.
What is the sum of the first 21 terms of the AP?
I. The common difference of the A.P. is 3
II. The $$11^{th}$$ term of the A.P. is 31.
Solution
If a is first term and d is common difference of an AP.
Then nth term of that AP is$$=a+(n-1)d.$$
Sum of n terms of an AP is$$=\ \frac{\ n}{2}\left\{2a+\left(n-1\right)d\right\}.$$
So, from statement I wecan not say the value of a and n.
II.
11th term of that AP is 31,
So, $$a+\left(11-1\right)3=31.$$
or,a=1.
Sum of first 21 terms is $$=\ \frac{\ 21}{2}\left\{2\times1+\left(21-1\right)\times3\right\}.$$
$$\ =\frac{\ 21}{2}\left\{2+60\right\}.$$
$$\ =\ 21\times\ 31=651.$$
So, B is correct choice.
Is 10, a factor of n + 5 ?
I. n is odd and divisible by 9
II. n is even and divisible by 5
Solution
I.
Let say value of n is 9 which is divisible by 9.
So,$$n+5=9+5=14$$,10 is not a factor of 14.
But, if value of n is 45 which is also divisible by 9, gives $$n+5=45+5=50$$.
10 is a factor of 50.
So, from statement I ,we can not say that whether 10 is a factor n+5 or not.
II.
n is even and also divisible 5 , this statement is only possible when the last digit of the number is 0.
In that case 10 is never be a factor of n+5.
So, B is correct choice.
Is the positive integer m odd?
I. $$m^2 + 2m$$ is even
II. $$m^2 + m$$ is even
Solution
I.
let put $$m=1 and m=2$$
when $$m=1$$ ,$$m^2+2m=1^2+2=3.$$ (which is an odd number.)
when$$m=2$$ ,$$m^2+2m=2^2+4=8.$$(which is an even number.)
So we can say that $$m^2+2m$$ is only even when m is an even number.
II.
let put $$m=1 and m=2$$
when $$m=1$$ ,$$m^2+m=1^2+1=2.$$
when $$m=2$$ ,$$m^2+m=2^2+2=6.$$
both the numbers are even numbers. So we can not say that whether m is odd or even.
So, A is correct choice.
What is the value of $$\cos \theta$$?
I. $$\sec \theta + \tan \theta$$ = 5
II. $$1 + \sin \theta = \frac{25}{13}$$
Solution
I.
we know that $$\sec^2\theta\ -\tan^2\theta\ =1.$$
or,$$\left(\sec\theta\ -\tan\theta\ \right)\left(\sec\theta\ +\tan\theta\ \right)=1.$$
here,$$\sec\theta\ +\tan\theta\ =5.$$
So,$$\sec\theta\ -\tan\theta\ =1\div5=0.2.$$
So, $$2\sec\theta\ =5+0.2=5.2.$$
or,$$\sec\theta\ =2.6.$$
or,$$\cos\theta\ =\frac{\ 10}{26}=\frac{\ 5}{13}.$$
II.
we know that $$\sin^2\theta\ +\cos^2\theta\ =1.$$
or,$$\cos\theta\ =\pm\ \sqrt{1-\sin^2\theta\ \ }.$$
here,$$\sin\theta\ =\ \frac{\ 12}{13}.$$
So,$$\cos\theta\ =\pm\ \sqrt{1-\ \frac{\ 144}{169}\ }=\pm\ \sqrt{\ \frac{\ 25}{169}\ }=\ \pm\ \frac{\ 5}{13}.$$
So, Option A is correct choice.
If the A.M. of 4, 20, 40, x is m, then what is the value of m?
I. $$\frac{m}{x} - 5$$ = 8
II. x is less than m
Solution
Arithmetic mean of 4,20,40 and x is $$\left(4+20+40+x\right)\div4.$$
So,$$m=\left(64+x\right)\div4=\left(16+\ \frac{\ x}{4}\right).$$
I.
$$\ \frac{\ m}{x}=8+5=13.$$
or,$$\ \ m=13x.$$
So,$$\ 13x=16+\ \frac{\ x}{4}.$$
or,$$\ 51x=64.$$
or,$$\ x=\ \frac{\ 64}{51}.$$
So,$$\ m=13x=13\times\ \frac{\ 64}{51}.$$
II. x is less than m, does not signifies any value of m.
So,A is correct choice.
What is the volume of the sphere?
I. The surface area of the sphere is equal to area of the circle with radius 5 cm.
II. The radius of the sphere is equal to the perimeter of a square.
Solution
Surface area of sphere is$$=4\pi\ r^2$$.
I. $$r=5$$.
Area of sphere$$=4\pi\ \times\ 5^2=100\pi\ .$$
II.
Perimeter of square $$=4a.\left(a=length\ of\ side\right).$$
So, from statement II we cant say what is the value of radius of sphere.
So, Option A is correct.
What is the range of y?
I. $$13 \leq x + y \leq 19$$
II. $$4 \geq x - y \geq -5$$
Solution
$$13\le\ x+y\le\ 19$$ and $$-5\le\ x-y\le4$$
So, $$13+5\le\left(x+y\right)-\left(\ x-y\right)\le19-4.$$
or,$$9\ge\ y\ge\ 7.5.$$
So, both the equation needed.
C is correct choice.
What is the rate of compound imterest?
I. Principal is Rs. 1200
II. Amount is Rs. 1323
Solution
We know that if Principal amount of P is compounded in n years at the rate of r percent,
Compounded Amount,A$$=P\times\ \left(1+\ \frac{\ r}{100}\right)^n$$.
But ,here P and A is given . r and n is missing.
So, it is not possible to calculate rate of interest.
D is correct choice.
For the Assertion (A) and Reason (R) the correct alternative from the following
(A) : The sum 1 + 3 + 5 + .... + 21 = 100
(R) : The sum of first $$n$$ odd positive integers is $$n^2$$
Solution
We know that the nth odd number is 2n−1 and the nth even number is 2n
$$Now, let's assume the sum of first n odd numbers to be S i.e. S=1+3+5+…+(2n−1)$$
Now let us add 1 n times to the right side,
$$S+n=(1+1)+(3+1)+(5+1)+…+(2n−1+1)$$
$$or,S+n=2+4+6+…+2n$$
Now adding these 2 equations, we get,
$$2S+n=1+2+3+…+(2n−1)+2n$$
The RHS is the sum of first 2n natural numbers which is as below,
$$2S+n=2n(2n+1)/2$$
$$or,2S+n=2n^2+n$$
$$or,2S=2n^2$$
$$or,S=n^2$$
$$So the sum of first n odd integers is n^2.$$
So, Condition R is true.
But Condition A is not true.
So, Option D is correct choice.
(A) : If the perimeter of a rectangle is 12 meters, then its maximum area is 9 $$m^2$$.
(R) : Geometric mean of two positive numbers is less than or equal to their Arithmetic mean
Solution
Geometric mean of numbers a and b is
$$=√ab.$$ and Arithmetic mean
$$=(a+b)/2$$.
we know that ,Geometric mean of two positive numbers is less than or equal to their Arithmetic mean.
So, Condition R is correct.
Condition A:
Let say ,length and breadth of the rectangle are l & b respectively.
So, $$2(l+b)=12.$$
Or,$$(l+b)=6.$$
And Area $$=lb$$.
multiplication of two numbers achieve when they are both equal in equation 1.
So,$$l=b=3.$$
Then only it can attain maximum value of
$$3×3=9$$m^2.
So, condition 'A' is true and correct explanation made by 'R'.
So,Option A is correct choice.
(A): If D, E and F respectively represent orthocentre, centroid and circumcentre of a triangle $$\triangle ABC$$, then Area of $$\triangle DEF = \frac{1}{4}$$ (Area of $$\triangle ABC$$)
(R) : In any triangle, orthocentre, centroid and circumcentre are collinear:
Solution
The centroid divides the distance from the orthocentre to the circumcentre in the ratio 2:1. So, they are colinear.
So, Condition R is true.
But, Condition A doesn't true as they are colinear.So, they can't form any triangle.
So,Option D is correct choice.
(A) : The sum of all internal angles in a pentagon is 540°
(R) : A pentagon has 5 sides
Solution
Sum of all internal of a polygon is
$$(n-2)180°.$$
Condition R:
we know that a pentagon has 5 sides.
So, Condition A:
Sum of all internal angles of a pentagon is
$$(5-2)×180°=3×180°=540°.$$
So, both the conditions are correct and A is dependent on R.
So, Option A is correct choice.
(A): The area of an equilateral triangle of side 5 cm is $$\frac{25}{4}\sqrt{3} cm^2$$
(R): Sum of the angles in the triangle is 180°
Solution
If side of an equilateral triangle is 'a' then the height of the triangle is$$=(√3/2)a.$$
Area of the triangle$$=(√3/4)a^2.$$
Condition A:
here ,$$a=5$$
so, Area$$=(√3/4)×5^2=(25/4)√3.$$
Condition R:
we know that sum of all the angle of triangle is 180°.
So, both the conditions are correct but 'A' is not dependent on 'R'.
Option B is correct choice.
(A): $$1^2 + 2^2 + 3^2 + ...... +49^2$$ = 40425
(R): Sum of the squares of the first n natural numbers is $$\frac{n(n + 1)(n + 2)}{6}$$
Solution
We know that sum of squares of first n naturals numbers is $$n(n+1)(2n+1)/6$$.
Condition A:
$$n=49.$$
so,$$1^2+2^2+3^2+......+49^2$$
$$=49×(49+1)(49×2+1)/6$$
$$=49×50×99/6$$
$$=40425.$$
Condition R:
$$n(n+1)(n+2)/6$$
$$=49×50×51/6$$
$$=20825.$$
which doesn't satisfy the given equation.
So,Option C is correct choice.
Identify the odd thing from the following options.
Solution
Earth , Mercury and Venus all are planet except Moon
Solution
Awareness, Lore and Cadence is a form of knowledge about a particular subject
Goofy means foolish
Solution
Asparagus is a small plant
Rest all are the different types of fruit
Solution
Elevation
-the action or fact of raising or being raised to a higher or more important level, state, or position.
Rest all are synonyms of improvement
Solution
S A A R C
The SAARC is an economic and geopolitical organization of eight countries which are located in South Asia.
Rest all are the agencies
Choose the option containing the odd pair.
Solution
Deer : Fawn
Fawn - baby Deer
Horse : Foal
Foal - baby horse
Swan : Cygnet
cygnet - Baby Swan
Boar : sow
Sow - adult female pig
Solution
Bees : Apiculture
Study of Bees
Birds : Aviculture
Study of Birds
Fish : Pisciculture
Study of Fish
so,
Paddy : Horticulture
Study of plant physiology and propagation
For the following questions answer them individually
Mr. X starts from his office and goes 5 km towards East, then he turns left and moves 4 km again turning left and reaches his home after walking 8 km. Then the shortest distance between his home and office, is (in kilometers)
Solution
shortest distace,
$$4^2 + 3^2 = 25 = 5^2$$
= 5 km
When the time is 1.15 P.m the minutes hand in the clock points towards North, then at 830 a.m. on the next day,the direction of minutes hand shows is :
Solution
East
A is to the north of B, C is to the east of B, then the direction of A with respect to C is
Solution
The direction of A with respect to C is North - West
Based on the information below answer the questions.
In a certain code, the 26 letters of english alphabets are written around a circle in the same order and each consonant is coded as the 4th consonant after it and each vowel is coded as the 3rd vowel after it.
The code for the word 'PRINCE' is
The word which is coded as KISJ is
Solution
K 11.....is coded as the 4th consonant before it, F = 6
I 9.... is coded as the 3rd vowel before it U = 21
S 19......is coded as the 4th consonant before it N = 14
J 10.....is coded as the 4th consonant before it D = 4
The letter 'X' is coded as
Solution
X = 24.....is coded as the 4th consonant after it is C =3
The letter that is coded as 'L' is
Solution
G ..... is coded as the 4th consonant after it is L
The word BOND is coded as
Solution
B 2 ......the 4th consonant after it is G
O 15....each vowel is coded as the 3rd vowel after it E
N 12 ......the 4th consonant after it is S
D 4 ......the 4th consonant after it is J
so,
GESJ
For the following questions answer them individually
If the word LINE is coded as QMQG then, in the same code, the word FORT is coded as
Solution
L...... after 4 alphabets Q
I...... after 3 alphabets M
N...... after 2 alphabets Q
E...... after 1 alphabets G
similarly
F...... after 4 alphabets k
O...... after 3 alphabets S
R...... after 2 alphabets U
T...... after 1 alphabets V
If the word BRING is coded as 50 and STATE is coded as 65 then in the same code, the word SWORD is coded as
Solution
Addition of place value
BRING = 2+18+9+ 14+ 7 = 48
STATE = 19+20+1+20+5 = 65
so,
SWORD = 19+23+15+18 + 4 = 79
In a certain code the word 'PRINCE' • coded as QQ JM DD. Then in the same code,CHARIT is coded as
Solution
'PRINCE' • coded as QQ JM DD
Nearest alphabet like
P Q R = QQ (nearest letter)
I J K L M N= J M (nearest letter)
C D E = DD (nearest letter)
similarly for CHARIT
C D E F G H = D G (nearest letter)
A B C D E F G H I J K L M N O P Q R =B Q (nearest letter)
I J K L M N O P Q R S T =J S (nearest letter)
D G B Q J S
In a certain code, the word NESTUM is coded as 123456 and the word PARIS is coded as 78903, then in the same code, the code for the word TAMPER is
Solution
N 1
E 2
S 3
T 4
U 5
M 6
P 7
A 8
R 9
I 0
S 3
so,
T = 4
A = 8
M = 6
P = 7
E = 2
R = 3
486729
In a certain code, the word MISTER is coded as SIYTKR, then the word NORMAL is coded as
Solution
MISTER is coded as SIYTKR
M.... after 5 alphabets s
I..... same I
S.... after 5 alphabets Y
T...same Y
E.... after 5 alphabets K
R ....same R
similarly for
N.... after 5 alphabets T
O.....O
R.... after 5 alphabets X
M..... M
A.... after 5 alphabets G
L.... l
TOXMGL
P is mother of V; V is sister of B. A is son of B. D is brother of A. S is mother of D. G is grand daughter of P. T has only two children V and B. Which one of the following is true?
Solution
we dont know the exact relation of G
so,
D is a son of T's son
In a family D is the wife of C and daughter of F and D is also the daughter-in-law of A. G is the mother of C. A is the paternal grand son of E. Then how A is related to G
Read the following information answer the questions below.
The following are the conditions for the selection of a manager post in an organisation.
i) The candidate must be a graduate with at least 65% of marks.
ii) As on 1st January 2018, the candidate's age must be between 22 and 30 years.
iii) The candidate must secure at least 40% marks in the entrance exam and at least 50% marks in the interview.
iv) The candidate must have management trainee experience of atleast 1 year.
(a) In case a candidate fails to fulfil (i) but has a post graduate degree with atleast 60% of marks, then his/her case is to be referred to Head of human resource section.
(b) In case a candidate fails to fulfil (iv) above but ready to execute a bond for two years then his/her case is to be referred to the General Manager.
Lakshmi was born on $$15^{th}$$ March 1993. She did her MBA and securecie marks. At a company she had a training experience after her MBA, for 13 months. Every member in the interview board gave marks from 6 to 9 out of 10. She secured 52% of marks in the entrance examination. Then
Solution
The data provided is not adequate to take a decision :
we don't know the graduation and post graduation percentage
Amar is an applicant for manager post and was born on 2nd September 1990. In the entrance , examination he got more marks than in interview. He got 55% marks in interview. He is ready to execute a bond for two years. He got a first class in graduation securing 68% of marks.
Solution
Arnar's case is to be referred to General Manager because
His age is 28
he qualified interview and entrance exam with required percentage
He got a first class in graduation securing 68% of marks.
He is ready to execute a bond for two years.
all the condition are fulfilled by Arnar
Sobhan graduated with 55% marks and secured post graduate degree by scoring 63% marks. In the entrance exam and interview he secured respectively 43% and 53% marks. He had experience as management trainee for 2 years. Sobhan's date of birth is 15th December 1986.
Solution
Sobhan is not selected for the post;
His graduation marks is less than 60%
Based on the below information, answer the questions.
5 persons A, B, C, D and E are to be invited on to the dais and are requested to sit on the 5 coloured chairs arranged in the order of Red, Blue, Green, Yellow and White. The person D is not allowed to sit on the Blue chair. The Chief guest A is to be seated on the white chair only. The person E is always to be seated by the side of A. The person B always to be seated at the middle.
The order in which the person are to be seated is:
Solution
so order is
D
C
B
E
A
If all the persons change their seats except the person sitting on the white chair, the person E is to be seated in the middle chair and no one has the same earlier neighbours, then the order of the persons to be seated on the dais is
Solution
so order is
D
C
B
E
A
Above is the original seating arrangement
If all the persons change their seats except the person sitting on the white chair, the person E is to be seated in the middle chair and no one has the same earlier neighbours
so,
B , D,E,C,A
For the following questions answer them individually
Observe the following figure.
The minimum number of cubes to be placed in empty spaces to complete the cube is
A hollow cuboid is constructed with identical small blocks as shown in the figure. Three such identical cuboids are attached to it to form z a big hollow cuboid. Then the number of small blocks required to fill the hollow space is
Seven letters G, N, T, Y, V, Q, C are selected and arranged in the dictionary order. Another two letters are selected and placed in between these seven letters so that
i) the position of the middle letter is not altered after the inclusion of these two letters.
ii) one of the two selected letters become the middle one for the right most five letters.
iii) for the left most five letters G is the middle letter and no two letters are consecutive. Then the two letters included are
Consider the following activities in a student life.
A: Examinations
B : Convocation
C : Admission
D : Results
E : First Class
If the sequence of occurrance of these activities is arranged in the reverse order, then the correct one in the given options is
Solution
A: Examinations
B : Convocation
C : Admission
D : Results
E : First Class
So the main order will be first his/her admission will be done, then examinations will be conducted followed by his/her results in a class (First, second, third etc) and finally degree will be awarded which is convocation.
Hence the correct order is CADEB
Then reverse order is
B : Convocation
E : First Class
D : Results
A: Examinations
C : Admission
In a class of 50 students, all students are ranked as per their marks in an examination. All the 50 students got distinct marks from each other. A student Ram is in $$18^{th}$$ position from the top and another student Syam is in $$26^{th}$$ position from top. The position of the student from the bottom who got the rank exactly in between the ranks of Ram and Syam, is
Solution
The position of the student from the bottom who got the rank exactly in between the ranks of Ram and Syam, is
29th rank from bottom
A, B, C, D and E are 5 boys sitting in a row. B is on the right of C who is on the right of A. The boy E is only one in between A and C. D is on the extreem left side. Then the boy in the middle of the arrangement is
Solution
E is in middle
