# SSC CPO Quantitative Aptitude Quiz 19

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# SSC CPO Quantitative Aptitude Quiz 19

### Introduction

What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.

A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.

The article SSC CPO Quantitative Aptitude Quiz 19 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. The article SSC CPO Quantitative Aptitude Quiz 19 will assist the students to know the expected questions from Quantitative Aptitude.

### Quiz

1. Find the largest number of 4-digits divisible by 12, 15 and 18.

A. 9900
B. 9750
C. 9450
D. 9000

Explanation:

Largest 4-digit number is 9999.
Remember: The question may be asked in a tricky way. Here, the largest number does not mean H.C.F.. We have to find a number which is divisible by 12, 15 and 18
The required largest number must be divisible by the L.C.M. of 12, 15 and 18
L.C.M. of 12, 15 and 18
12 = 2 × 2 × 3
15 =5 × 3
18 = 2 × 3 × 3
L.C.M. = 180
Now divide 9999 by 180, we get remainder as 99
The required largest number = (9999 – 99) =9900
Number 9900 is exactly divisible by 180

2. H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.

A. 230, 140
B. 215, 130
C. 195, 143
D. 155, 115

Explanation:

Hint: Product of two numbers = Product of their H.C.F. and L.C.M.
Given:
1) H.C.F. of two numbers = 13
2) The numbers are in the ratio of 15 : 11

Let the two numbers be 15y and 11y
H.C.F. is the product of common factors
Therefore, H.C.F. is y. So y = 13

The two numbers are:
15y = 15 × 13 = 195
11y = 11 × 13 = 143
We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)
Product of H.C.F. and L.C.M. = 13 × 2145 = 27885
Product of two numbers = 195 × 143 = 27885
Hence, the calculated answer is correct.

3. If the product and H.C.F. of two numbers are 4107 and 37 respectively, then find the greater number.

A. 111
B. 222
C. 332
D. 452

Explanation:

4107 is the square of 37.
So let two numbers be 37x and 37y.
37x × 37y = 4107
xy = 3
3 is the product of (1 and 3)
x = 1 and y = 3
37x = 37 × 1 =37
37y = 37 × 3 = 111
Greater number = 111

4. Find the sum of two numbers, which are greater than 29 and have H.C.F. and L.C.M. of 29 and 4147 respectively.

A. 858
B. 696
C. 1050
D. 4147

Explanation:

Product of two numbers = Product of their H.C.F. and L.C.M.
Product of two numbers = 29 × 4147 = 120263
Given: Two numbers are greater than 29.
Therefore, let the two numbers be 29x and 29y.
So, 29x × 29y = 120263
Xy = 143
Factors of 143 are: 1, 11, 13, and 143

Case: 1) If we consider factors of 143 as 1 and 143 (co-primes), then we get the value of two numbers x and y = (29 and 4147) —— (Which is wrong: As it is given that, the numbers are greater than 29)
Case: 2) If we consider factors of 143 as 11 and 13 (co-primes), then we get the value of two numbers x and y = (319, 377) —— (These two values are greater than 2. So, it is the correct answer)
Therefore, the two numbers are 319 and 377.
Sum of two numbers = 319 + 377 = 696

5. Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

A. 5 : 28 : 00 hrs
B. 5 : 30: 00 hrs

C. 5: 38 : 00 hrs

D. 5 : 40: 00 hrs

Explanation:
The number on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Hence, make the dividend completely divisible by the divisor. This is possible if we subtract the remainder from the dividend.
Therefore,
1657 – 6 = 1651
2037 – 5 = 2032
H.C.F. of 1651 and 2032 is 127. 127 is the common factor.
127 × 13 = 1651
Thus by adding 6, we get 1651 + 6 = 1657

1. The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00 hours, then find the time at which they will change simultaneously.

A. 5 : 28 : 00 hrs
B. 5 : 30 : 00 hrs

C. 5 : 38 : 00 hrs
D. 5 : 40 : 00 hrs

Explanation:

Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 : 00 hrs.

2. A rectangular courtyard 4.55 meters long and 5.25 meters wide is paved exactly with square tiles of the same size. Find the largest size of the tile used for this purpose?

A. 25 cm
B. 21 cm
C. 45 cm

D. 35 cm

Explanation:
Here, we are asked to find the largest size of tile. Therefore, calculate H.C.F.
Step 1: Covert numbers without decimal places i.e 455 cm and 525 cm
Step 2: Find the H.C.F. of 455 and 525
H.C.F. of 455 and 525 = 35 cm
Hence, the largest size of the tile is 35 cm.

3. John, Smith, and Kate start at the same time, the same point and in the same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?

A. 30 min
B. 25 min

C. 20 min
D. 15 min

Explanation:
L.C.M. of 250, 300 and 150 = 1500 sec
Dividing 1500 by 60 we get 25, which mean 25 minutes.
John, Smith and Kate meet after 25 minutes.

4. If Suresh distributes his pens in the ratio of $$\frac{1}{2 }$$ : $$\frac{1}{4 }$$ : $$\frac{1}{5 }$$ : $$\frac{1}{7 }$$ between his four friends A, B, C and D, then find the total number of pens Suresh should have?

A. 153
B. 150
C. 100
D. 125

Explanation:
Here, A : B : C : D = $$\frac{1}{2 }$$ :$$\frac{1}{4 }$$ : $$\frac{1}{5 }$$ : $$\frac{1}{7 }$$

1) L.C.M of 2, 4, 5, 7 is 140
2) Find the number of pens each friend received ——— (To find no. of pens each friend has, multiply the ratio with the L.C.M. calculated)

A = ($$\frac{1}{2 }$$) x 140 = 70
B = ($$\frac{1}{4 }$$) x 140 = 35
C = ($$\frac{1}{5 }$$) x 140 = 28
D = ($$\frac{1}{7 }$$) x 140 = 20

3) Total number of pens = (70 x + 35 x + 28 x + 20 x) = 153 x
Minimum number of pens (x) = 1
Therefore, total number of pens = 153 pens.

5. A man, his wife, and daughter worked in a garden. The man worked for 3 days, his wife for 2 days and daughter for 4 days. The ratio of daily wages for man to women is 5 : 4 and the ratio for man to daughter is 5 : 3. If their total earnings are mounted to Rs. 105, then find the daily wage of the daughter.

A. Rs. 15
B. Rs. 12
C. Rs. 10
D.

Explanation:
Assume that the daily wages of man, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.
Multiply (no. of days) with (assumed daily wage) of each person to calculate the value of x.
[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 105
[15x + 8x + 12x] = 105
35x = 105
x = 3
Hence, man’s daily wage = 5x = 5 x 3 = Rs. 15
Wife’s daily wage = 4x = 4 x 3 = Rs. 12
Daughter’s daily wage = 3x = 3 x 3 = Rs. 9

1. Amit, Raju and Ram agree to pay their total electricity bill in the proportion 3 : 4 : 5. Amit pays first day’s bill of Rs. 50, Raju pays second day’s bill of Rs. 55 and Ram pays third day’s bill of Rs. 75. How much amount should Amit pay to settle the accounts?

A. Rs. 15.25

B. Rs. 17
C. Rs. 12

D. Rs. 5

Explanation:
Toatal bill paid by Amit, Raju and Ram = ( 50 + 55 +75 ) = Rs. 180
Let amount paid by Amit, Raju and Ram be Rs. 3x, 4x and 5x respectively.
Therefore, (3x + 4x + 5x ) = 180
12x = 180
x = 15
Therefore, the amount paid by,
Amit = Rs. 45
Raju = Rs. 60
Ram = Rs. 75
But actually as given in the question, Amit pays Rs. 50, Raju pays Rs. 55 and Ram pays Rs. 80. Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he needs to pay Rs. 5 to Raju settle the amount.

2. If the length is increased by 25%, by what percent the width of a rectangle should be decreased so as to keep the area same.

A. 25%
B. 20%
C. 30%

D. 10%

Explanation:
Let the original length be l and the width be b
Therefore, the area = l*b
Now, as the length is increased by 25%, the new length is (1.25*l) and let the new width be x.
As the area is same, 1.25*l*x = l*b
x = b/1.25 = 0.8b
Therefore, the width is to be decreased by 20%.

3. A bag contains an equal number of 25 paise, 50 paise, and one rupee coins respectively. If the total value is Rs 105, how many types of each type are present?

A. 75 coins
B. 60 coins
C. 30 coins
D. 25 coins

Explanation:

Bag consists of 25 paise, 50 paise and 1 rupee (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4

Total value of 25 paise coins = ($$\frac{1}{7 }$$ ) x 105 = 15
Total value of 50 paise coins = ($$\frac{2}{7 }$$) x 105 = 30
Total value of 100 paise coins = ($$\frac{4}{7 }$$) x 105 = 60

No. of 25 paise coins = 15 x 4 = 60 coins
No. of 50 paise coins = 30 x 2 = 60 coins
No. of 1 rupee coins = 60 x 1 = 60 coins

4. A purse contains 342 coins consisting of one rupees, 50 paise and 25 paise coins. If their values are in the ratio of 11 : 9 : 5 then find the number of 50 paise coins?

A. 180
B. 150
C. 162
D. 99

Explanation:

Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.

No. of 1 rupee coins = $$\frac{11x }{1 }$$ = 11x
No. of 50 paise coins = $$\frac{9x }{0.5 }$$ = 18x
No. of 25 paise coins = $$\frac{15x }{0.25 }$$ = 20x

11x + 18x + 9x = 342
38x = 342
x = 9

Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins
No. of 50 paise coins = 18 x 9 = 162 coins
No. of 25 paise coins = 20 x 9 = 180 coins

5. When folded into two equal halves a rectangular sheet had a perimeter of 48cm for each part folded along one set of sides and the same is 66cm when folded along the other set of sides. Find the area of the sheet.

1584
B. 1120
C. 792
D. 1320

Explanation:

Let the sheet be folded along its breadth and its perimeter = 48cm
Therefore, ($$\frac{l }{2 }$$ + b) = 48 …… (i)
Now, let the sheet be folded along its length, and the perimeter = 66cm
(l + $$\frac{b }{2 }$$)= 66 …… (ii)
Solving (i) and (ii), we get,
l = 56cm, b = 20cm
Area = l*b
Area = 1120 cm2