SBI Clerk Expected Quant Questions Set-2
Download SBI Clerk Expected Quant Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Expected Quant questions with solutions.
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Question 1: In 8 years, Ramesh will be twice as old as Farheen was 8 years ago. If Ramesh is now 16 years older than Farheen, the present age of Farheen is :
a) 32
b) 40
c) 36
d) 44
e) 24
Question 2: A was thrice as old as B when C was 2 years old. A will be twice as old as B when C is 10 years old. How old will B be when the difference between the ages of A and B will be equal to the age of C?
a) 16 years
b) 22 years
c) 18 years
d) 24 years
e) 32 years
Question 3: The ratio of P’s age 3 years ago and Q’s age after 5 years is 6 : 13. If at present, the ratio of their ages is 11 : 20, then what is the ratio of P’s age 2 years hence and Q’s age 5 years ago?
a) $\frac{9}{11}$
b) $\frac{7}{9}$
c) $\frac{5}{13}$
d) $\frac{7}{11}$
e) $\frac{8}{13}$
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Question 4: The sum of the ages of a P and Q is 53 years. Four years ago, the product of their ages was two times the square of Q’s age. The present age of P is?
a) 34 years
b) 38 years
c) 32 years
d) 42 years
e) 36 years
Question 5: The present age of Sanjay is 40 percent of his mother’s present age. 8 years from now, he will be half as old as his mother. What is the sum of present ages of Sanjay and his mother?
a) 42 years
b) 70 years
c) 56 years
d) 84 years
e) 50 years
Question 6: Find the approximate value of $\Large\frac{169.87^{2}-13.94^{2}}{183.87}$ $\large+$ $\Large\frac{3211.96}{44.002}$
a) 224
b) 236
c) 239
d) 229
e) None of these
Question 7: Find the approximate value of $17.33+142.895-76.795+235.008-4.779+38.102$
a) 321
b) 351
c) 281
d) 371
e) None of these
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Question 8: Find the approximate value of ‘x’ in $\Large\frac{383.96}{x} = \frac{x}{24.01}$
a) 84
b) 96
c) 92
d) 88
e) None of these
Question 9: Find the approximate value of $\sqrt{6560}\div2.98^{2}+\sqrt[3]{2189}\div\sqrt{2700}\times\sqrt{5180}\div\sqrt[3]{25}$
a) 11
b) 15
c) 20
d) 5
e) None of these
Question 10: Find the approximate value of $153.001\times146.95-66.009\times73.952+25.001\times23.856$
a) 17952
b) 18207
c) 18456
d) 17605
e) None of these
Question 11: The average marks of a class increased by $\large\frac{1}{3}$ when a student’s marks were wrongly entered as 64 instead of 46. The number of students in the class are
a) 54
b) 58
c) 63
d) 57
e) None of these
Question 12: The average age of students in a class is 12. If the average age of boys is 18 and the average age of girls is 14 and If the number of boys are 22, then find the total number of students in the class
a) 52
b) 55
c) 60
d) 58
e) None of these
Question 13: In the first 20 overs of a cricket game, the run rate was 4.5, What should be the run rate in the next 30 overs of the game to chase the target of 210?
a) 6
b) 4
c) 8
d) 6.5
e) None of these
Question 14: In a society, 60% of residents own a car and 40% own a bike. Nobody owns car and bike both. Also, 25% of the residents work in the banking sector. If 20% of those who work in the banking sector own a bike, then the residents who own a car and work in non-banking sector form what percentage of the total people?
a) 25%
b) 40%
c) 37.5%
d) 60%
e) 42.5%
Question 15: Amit scored 10% less marks than Sumit and 20% more marks than Rakesh. If Tina scored 35 marks more than Amit and Sumit scored 20 marks less than Tina, what is the absolute difference between the marks of Tina and Rakesh?
a) 75
b) 82.5
c) 37.5
d) 57.5
e) 40
Instructions
Select the correct option.
Question 16: Quantity 1:A number is increased by 85% and then it is decreased by 15% and result is approximately 357
Quantity 2:A number is increased by 82% and then it is decreased by 12% and the result is approximately 355.5
a) Quantity 1 $\leq$Quantity 2
b) Quantity 1 $\geq$Quantity 2
c) Quantity 1 < Quantity 2
d) Quantity 1 > Quantity 2
e) Quantity 1 = Quantity 2
Question 17: Quantity 1:If the cost price of total apples bought is Rs 288 and each apple is sold at Rs 14 each and the profit percent is 50/3 then number of apples sold is
Quantity 2:An article is marked up by 40% an a discount of 30% was given then 13 times the loss percent is
a) Quantity 1 $\leq$Quantity 2
b) Quantity 1 $\geq$Quantity 2
c) Quantity 1 < Quantity 2
d) Quantity 1 > Quantity 2
e) Quantity 1 = Quantity 2
Question 18: Quantity 1:Numerator of a fraction is increased by 20% an the denominator is decreased by 25% and the resultant is 2 then the sum of numerator and denominator of the original fraction.
Quantity 2:Numerator of a fraction is increased by 50% and denominator is decreased by 60% and the resultant is 3 then the sum of numerator and denominator of original fraction is
a) Quantity 1 $\leq$Quantity 2
b) Quantity 1 $\geq$Quantity 2
c) Quantity 1 < Quantity 2
d) Quantity 1 > Quantity 2
e) Quantity 1 = Quantity 2
Question 19: Quantity 1:Roots of the equation $6x^{2}-13x+6$=0
Quantity 2:Roots of the equation $25x^{2}-25x+6$=0
a) Quantity 1 $\leq$Quantity 2
b) Quantity 1 $\geq$Quantity 2
c) Quantity 1 < Quantity 2
d) Quantity 1 > Quantity 2
e) Quantity 1 = Quantity 2
Instructions
Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.
Question 20: Quantity I: In a party, each person shakes hand with every other person. If there are total 45 handshakes, then the total number of persons present are?
Quantity II: In a party, each person shakes hand with every other person. If there are 8 persons, then the number of handshakes possible are?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I $\geq$ Quantity II
d) Quantity I $\leq$ Quantity II
e) Quantity I = Quantity II or The relationship between Quantity I and Quantity II cannot be established.
General Knowledge Questions & Answers PDF
Answers & Solutions:
1) Answer (B)
Let us assume the present age of Farheen = x years
So present age of Ramesh will be = x + 16 years
Ramesh’s age after 8 years will be = x+16+8 = x+24 years
Farheen’s age 8 years ago = x – 8 years
$\Rightarrow$ x+24 = 2*(x-8)
$\Rightarrow$ x= 40 years
So the present age of Farheen will be 40 years.
2) Answer (B)
Let the age of A and B initially be ‘a’ and ‘b’ respectively.
When C = 2, a/b = 3
=> a = 3b
When C = 10, 8 years have passed.
=> (a+8)/(b+8) = 2
a+8 = 2b + 16
a = 2b+8
=> b = 8 years and a = 24 years.
When C is as old as the difference between the ages of A and B, i.e, when C is 24-8 = 16 years old, age of B will be 8+14 = 22 years. Therefore, option B is the right answer.
3) Answer (D)
We have,
$\frac{P – 3}{Q + 5} = \frac{6}{13}$
So, 13P – 39 = 6Q + 30
13P – 6Q = 69
$\frac{P}{Q} = \frac{11}{20}$
20P – 11Q = 0
Solving both the equations we have,
P = 33 years and Q = 60 years
So, $\frac{P + 2}{Q – 5} = \frac{35}{55} = \frac{7}{11}$
Hence, option D is the right answer.
4) Answer (A)
Let the age of P and Q be p and q respectively.
We have, p+q = 53
$(p-4)(q-4) = 2 * (q-4)^2$
p-4 = 2q – 8
p = 2q – 4
Substituing the value of p in first equation,
3q – 4 = 53
q = 19
p = 53 – 19 = 34
Hence, option A is right choice.
5) Answer (C)
Let the present age of Sanjay’s mother be ‘x’. So Sanjay’s present age will be .4x
We have been given that
(.4x + 8)*2 = x + 8
=> .8x + 16 = x + 8
=> .2x = 8
=> x = 40
Hence, the present age of the mother is 40 years. Thus, required sum = 56.
6) Answer (D)
$\Large\frac{169.87^{2}-13.94^{2}}{183.87}$ $\large+$ $\Large\frac{3211.96}{44.002}\simeq \frac{170^{2}-14^{2}}{184}$ $\large+$ $\Large\frac{3212}{44}$
$= \Large\frac{(170+14)(170-14)}{184}$ $+$ $73$
$= \Large\frac{184\times156}{184}$ $+$ $73$
$= 156+73 = 229$
7) Answer (B)
$17.33+142.895-76.795+235.008-4.779+38.102 \simeq 17+143-77+235-5+38$
Adding all positive terms:
$17+143+235+38 = 433$
Adding all negative terms:
$77+5 = 82$
$\therefore$ $17+143-77+235-5+38 = 433-82 = 351$
8) Answer (B)
$\Large\frac{383.96}{x} = \frac{x}{24.01}$
$\Rightarrow x^{2} \simeq 384\times24 = 9216$
$\therefore$ x $= \sqrt{9216} = 96$
9) Answer (B)
$\sqrt{6560}\div2.98^{2}+\sqrt[3]{2189}\div\sqrt{2700}\times\sqrt{5180}\div\sqrt[3]{25} \simeq \sqrt{6561}\div3^{2}+\sqrt[3]{2197}\div\sqrt{2704}\times\sqrt{5184}\div\sqrt[3]{27}$
$= \Large\frac{\sqrt{6561}}{3^{2}}+\frac{\sqrt[3]{2197}\times\sqrt{5184}}{\sqrt{2704}\times\sqrt[3]{27}}$
$= \Large\frac{81}{9}+\frac{13\times72}{52\times3}$ $= 9+6 = 15$
10) Answer (B)
$153.001\times146.95-66.009\times73.952+25.001\times23.856 \simeq 153\times147-66\times74+25\times24$
$153\times147$ can be written as $(150+3)(150-3)$
$=$ $150^{2}-3^{2}$ $(\because (a+b)(a-b) = a^{2}-b^{2})$
$=$ $22500-9$ $=$ $22491$
$66\times74$ can be written as $(70-4)(70+4)$ $=$ $70^{2}-4^{2}$ $=$ $4900-16 = 4884$
$25\times24$ can be written as $24\times\Large\frac{100}{4} =$ $600$
$\therefore$ $153\times147-66\times74+25\times24 = 22491-4884+600 = 18207$
11) Answer (A)
Let the number of students be ‘x’
Given that the average is increased by $\Large\frac{1}{3}$ due to the increase of $18$ marks ( $64$ $-$ $46$ $=$ $18$ marks )
Actual increase in marks $=$ $\Large\frac{1}{3}$ $\times$ x
$\Rightarrow$ $\Large\frac{1}{3}$ $\times$ x $=$ $18$
$\therefore$ Number of students(x) $=$ $54$
12) Answer (B)
Let the total age of boys be ‘B’
Average age of boys = 18
$\Large\frac{B}{22}$ $=$ $18$
$\Rightarrow$ B $=$ $396$
Let the total age of Girls be ‘G’
Let the number of girls be ‘x’
Average age of girls $=$ $14$
$\Large\frac{G}{x}$ $=$ $14$
$\Rightarrow$ G $=$ $14$x
Total average of students $=$ $12$
Total age of students $=$ Total age of boys $+$ Total age of girls $=$ $396$ $+$ $14$x
Total number of students $=$ Number of boys $+$ Number of girls $=$ $22$ $+$ x
Average age of class $=$ $12$
$\Rightarrow$ $\Large\frac{396 + 14x}{22 + x}$ $=$ $12$
$\Rightarrow$ $396$ $+$ $14$x $=$ $462$ $+$ $12$x
$\Rightarrow$ $2$x $=$ $66$
$\Rightarrow$ x $=$ $33$
$\therefore$ Number of girls $=$ $33$
Total number of students $=$ $22$ $+$ $33$ $=$ $55$
13) Answer (B)
Run-rate = 4.5 runs per over
$1$ over $\rightarrow$ $4.5$ runs
$20$ overs $\rightarrow$ $20\times4.5$ $=$ $90$ runs
Total target $=$ $210$ runs
Remaining score $=$ $210$ $-$ $90$ $=$ $120$ runs
Remaining overs $=$ $30$ overs
$\therefore$ Required run-rate $=$ $\Large\frac{120}{30}$ $=$ $4$ runs per over
14) Answer (B)
Let the total number of residents be 100x
People who own car = 60k
People who own a bike = 40k
People who work in the banking sector = 25k
People in the banking sector who own a bike = 20% of 25k = 5k
Rest of the people who work in the banking sector must be owning a car
Number of people owning a car and working in banking sector = 25k – 5k = 20k
Number of people owning a car and working in non-banking sector = 60k – 20k = 40k
Required % = 40k/100k * 100 = 40%
Hence, option B is the correct answer.
15) Answer (D)
Let Sumit scored 100x marks
Then, Amit scored 90x marks
Amit’s score is 20% more than that of Rakesh.
So, Rakesh’s score = 75x
Tina’s score = Amit’s score + 35 = 90x + 35…..(i)
Tina’s score = Sumit’s score + 20 = 100x + 20…(ii)
On comparing (i) and (ii),
90x + 35 = 100x + 20
On solving, we get x = 1.5
So, Rakesh’s score = 75x = 75* 1.5 = 112.5
And Tina’s score = 90x + 35 = 170
Difference = (170 – 112.5) = 57.5
Hence, option D is the correct answer.
16) Answer (D)
In Q1 we have x*1.85*0.85=357
x=357/(1.85*0.85)
x=227
In Q2 we have x*1.82*0.88=355.5
x=355.5/(1.82*0.88)
x=220
Therefore Q1>Q2
17) Answer (C)
In Q1 we have Total CP=288
Let the number of apples be x
Total SP=14x
Therefore ((14x-288)/(288))*100=50/3
14x-288=48
14x=288+48
x=336/14
x=24
In Q2 we have CP=100
MP=100*1.4=140
SP=MP-discount
SP=140-140*(30/100)
SP=98
Loss percent=((100-98)/(100))*100
=2%
13 times the loss percent=26
Therefore Q2 > Q1
18) Answer (E)
In Q1 let the fraction be be (x/y)
Then as given (1.2x)/(0.75y)=2
x/y =5/4
sum=5+4=9
In Q2 let the fraction be (x/y)
Then as given (1.5x)/(0.4y)=3
x/y =4/5
sum=4+5=9
19) Answer (D)
Roots of the equation in Q1 is $6x^{2}-9x-4x+6$=0
3x(2x-3)-2(2x-3)=0
(3x-2)(2x-3)=0
x=2/3 and x=3/2
In Q2 we have $25x^{2}-15x-10x+6$=0
5x(5x-3)-2(5x-3)=0
(5x-2)(5x-3)=0
x=2/5 and x=3/5
And so by comparing Q1 > Q2
20) Answer (B)
Quantity I:
Number of handshakes possible = $\dfrac{n(n-1)}{2}$ where n = number of persons
Given, $\dfrac{n(n-1)}{2} = 45$
⇒ $n^2 – n = 90$
⇒ $n^1 – n – 90 = 0$
⇒ $n^2 – 10n + 9n – 90 = 0$
⇒ $n(n-10) + 9(n-10) = 0$
⇒ $(n-10)(n+9) = 0$
⇒ n = 10 or n = -9
Number of persons cannot be negative.
Hence, Total number of persons is 10.
Quantity II:
Number of handshakes possible = $\dfrac{n(n-1)}{2}$ where n = number of persons.
= $\dfrac{8 \times 7}{2} = 28$
Therefore, Quantity I < Quantity II
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