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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Jar A is sold at 18% profit & both Jars have the same cost price, if overall profit is 16%, and if Jar B is sold at Rs.70 more then find the cost price of Jar A.**Correct**Answer: 1) Rs. 500**

Explanation:

Let the cost price of each Jar is Rs. 100x

Total CP = 200x

Since overall profit is 16%

Sum of SP of both the jars = 116% of 200x = 232x

SP of jar A = 118% of 100x = 118x

SP of jar B = 232x – 118x = 114x

Since jar B is sold at Rs. 70 more

114x -100x = 14x

14x = 70

X = 5

Cost price of jar A = 100*5 = Rs. 500Incorrect**Answer: 1) Rs. 500**

Explanation:

Let the cost price of each Jar is Rs. 100x

Total CP = 200x

Since overall profit is 16%

Sum of SP of both the jars = 116% of 200x = 232x

SP of jar A = 118% of 100x = 118x

SP of jar B = 232x – 118x = 114x

Since jar B is sold at Rs. 70 more

114x -100x = 14x

14x = 70

X = 5

Cost price of jar A = 100*5 = Rs. 500 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**28 men can complete a work in 45 days, 35 men start work and work for x days after that 15 more men joined them and complete remaining work in 15(2/5) days, find the value of x.**Correct**Answer: 4) 14**

Explanation:

Let the total work is 28*45 = 1260

Work done by 35 men is x days = 35x

Work done by (35+15) men in 15(2/5) days = 50 * 77/5 = 770

Since the work is completed,

35x + 770 = 1260

X = 14Incorrect**Answer: 4) 14**

Explanation:

Let the total work is 28*45 = 1260

Work done by 35 men is x days = 35x

Work done by (35+15) men in 15(2/5) days = 50 * 77/5 = 770

Since the work is completed,

35x + 770 = 1260

X = 14 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**A mixture of milk and water has 30% water, 80 litre mixture is removed, then milk becomes 64 litres more than water in remaining mixture. Find initial milk.**Correct**Answer: 4) 168 litres**

Explanation:

Let the initial amount of milk in the mixture is 7x, and initial amount of water in the mixture is 3x

Now, 80 litres will be removed in the ratio as 7:3

Milk removed will be 56 litres and water removed will be 24 litres

Milk left = 7x-56, water left = 3x -24

According to question,

7x – 56 – (3x-24) = 64

4x – 32 = 64

X = 24

Therefore, initial milk = 7x = 7*24 = 168 litresIncorrect**Answer: 4) 168 litres**

Explanation:

Let the initial amount of milk in the mixture is 7x, and initial amount of water in the mixture is 3x

Now, 80 litres will be removed in the ratio as 7:3

Milk removed will be 56 litres and water removed will be 24 litres

Milk left = 7x-56, water left = 3x -24

According to question,

7x – 56 – (3x-24) = 64

4x – 32 = 64

X = 24

Therefore, initial milk = 7x = 7*24 = 168 litres - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**The ratio of the speed of a boat and current is 11:3. The ratio of time taken of distance “D” in downstream and “D+45” in upstream is 2: 5. Find the distance D.**Correct**Answer: 1) 105**

Explanation:

Let speed of boat = 11x and speed of current = 3x

Upstream speed =11x-3x =8x

Downstream speed = 11x+3x =14x

According to question,

D/14x : (D+45) / 8x = 2:5

D/14x * 8x/(D+45) = 2/5

20D = 14D + 14*45

D = 105Incorrect**Answer: 1) 105**

Explanation:

Let speed of boat = 11x and speed of current = 3x

Upstream speed =11x-3x =8x

Downstream speed = 11x+3x =14x

According to question,

D/14x : (D+45) / 8x = 2:5

D/14x * 8x/(D+45) = 2/5

20D = 14D + 14*45

D = 105 - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**A invests “K” and after 4 month B invest “K -1260″. The ratio of total annualprofit and B’s profit is 9: 2. Find the value of “K”.**Correct**Answer: 3) 2205**

Explanation:

Let the total annual profit = 9x

Profit of B = 2x

Profit of A will be = 9x-2x = 7x

(Investment of A * Time)/(Investment of B * Time ) = Profit of A / Profit of B

Investment of A is for 12 months and Investment of B is for (12-4) = 8 months

(K *12) / {(K-1260)*8} = 7/2

24K = 7*8* (K-1260)

3K = 7K – 7*1260

K = 2205Incorrect**Answer: 3) 2205**

Explanation:

Let the total annual profit = 9x

Profit of B = 2x

Profit of A will be = 9x-2x = 7x

(Investment of A * Time)/(Investment of B * Time ) = Profit of A / Profit of B

Investment of A is for 12 months and Investment of B is for (12-4) = 8 months

(K *12) / {(K-1260)*8} = 7/2

24K = 7*8* (K-1260)

3K = 7K – 7*1260

K = 2205 - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Ages of three friends A, B and C after 5 years will be in the ratio 8:9:6 and ratio of ages of B,C and D before 3 years was 7:4:5. What is the sum of the present age of B and D.**Correct**Answer: 4) 54 years**

Explanation:

Let the age of A,B and C after 5 years is 8x, 9x and 6x respectively.

Present age of A,B and C is 8x-5, 9x-5 and 6x-5 respectively.

Let the age of B, C and D before 3 years was 7y, 4y and 5y.

Present age of B, C and D is 7y +3, 4y +3, 5y + 3 .

Equating

the present age of B

9x-5 = 7y + 3

9x-7y = 8 …………..(1)

Equating the present age of C,

6x-5 = 4y + 3

6x – 4y = 8 ……….(2)

From, equation (1) and (2), we get

X = y = 4

Present age of B = 7y+3 = 7*4 + 3 = 31

Present age of D = 5y+3 = 5*4+3 = 23

Sum of the present age of B and D = 31+23 = 54 yearsIncorrect**Answer: 4) 54 years**

Explanation:

Let the age of A,B and C after 5 years is 8x, 9x and 6x respectively.

Present age of A,B and C is 8x-5, 9x-5 and 6x-5 respectively.

Let the age of B, C and D before 3 years was 7y, 4y and 5y.

Present age of B, C and D is 7y +3, 4y +3, 5y + 3 .

Equating

the present age of B

9x-5 = 7y + 3

9x-7y = 8 …………..(1)

Equating the present age of C,

6x-5 = 4y + 3

6x – 4y = 8 ……….(2)

From, equation (1) and (2), we get

X = y = 4

Present age of B = 7y+3 = 7*4 + 3 = 31

Present age of D = 5y+3 = 5*4+3 = 23

Sum of the present age of B and D = 31+23 = 54 years - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**When 60% of a number is added with that number and its result is multiplied by 15, then result becomes 480. Find the number?**Correct**Answer: 3) 20**:

Explanation

Let the number is x

60% of x = (60x/100) = 3x/5

According to question,

(x+3x/5)*15 = 480

(8x/5)*15 = 480

24x =480

X=20Incorrect**Answer: 3) 20**:

Explanation

Let the number is x

60% of x = (60x/100) = 3x/5

According to question,

(x+3x/5)*15 = 480

(8x/5)*15 = 480

24x =480

X=20 - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**If the average of three numbers A, B and C is 42, average of A and C is 39, average of B and C is 44 then what will be the 3/4th of the number C.**Correct**Answer: 1) 30**

Explanation:

(A+B+C)/3 = 42

(A+B+C) = 126 ……………(1)

(A+C)/2 = 39

(A+C) = 78 ………………(2)

Adding (A+C) = 78 , in equation (1), we get

B = 48

(B+C)/2 = 44

B+C = 88

48+C = 88

C =40

3/4th of the number C = (3/4)* 40 = 30Incorrect**Answer: 1) 30**

Explanation:

(A+B+C)/3 = 42

(A+B+C) = 126 ……………(1)

(A+C)/2 = 39

(A+C) = 78 ………………(2)

Adding (A+C) = 78 , in equation (1), we get

B = 48

(B+C)/2 = 44

B+C = 88

48+C = 88

C =40

3/4th of the number C = (3/4)* 40 = 30 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**The perimeter of a rectangle is 74 cm and its breadth is 5 cm less than length. An equilateral triangle has side same as the breadth of the rectangle. Area of the equilateral triangle is**Correct**Answer: 5) 64√3**

**Explanation:**

Perimeter of Rectangle = 2*(L+B)

2*(L+B) = 74

L+B = 37

Also, breadth is 5 less than length (L = B+5)

B+5+B = 37

2B = 32

B = 16

Area of the equilateral triangle = ( √3 /4) * (side)^{2}

= (√3 /4) * (16)^{2}

= 64 √3 sq cmIncorrect**Answer: 5) 64√3**

**Explanation:**

Perimeter of Rectangle = 2*(L+B)

2*(L+B) = 74

L+B = 37

Also, breadth is 5 less than length (L = B+5)

B+5+B = 37

2B = 32

B = 16

Area of the equilateral triangle = ( √3 /4) * (side)^{2}

= (√3 /4) * (16)^{2}

= 64 √3 sq cm - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude**Ajay invests a principal amount of Rs. 5600 in a fixed deposit for 3.5 years. The rate at which he invests is 16% per annum. What will be the simple interest earned by him at the end of his investment tenure?**Correct**Answer: 3) Rs. 3136**

Explanation:

Simple interest = (Principal*Rate*Time)/100

= (5600*16*3.5)/100

= 3136Incorrect**Answer: 3) Rs. 3136**

Explanation:

Simple interest = (Principal*Rate*Time)/100

= (5600*16*3.5)/100

= 3136

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