# PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

The sum of the present ages of a woman and her daughter is 45 years. When the woman attains her husband’s present age, the ratio of the ages of her husband and her daughter will be 2 : 1. Find the present age (in years) of her daughter.

- (a)
18

- (b)
15

- (c)
20

- (d)
12

Answer: Option B

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**Explanation** :

Let the present ages of the woman, her husband and her daughter be w years, h years and d years respectively.

w + d = 40 …(1)

The woman would attain her husband's age after (h - w) years.

∴ (h + h - w)/(d + h - w) = 2

w = 2d …(2)

From (1) and (2), d = 15

Hence, option (b).

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**Directions for next 2 questions : These questions are based on the information given below.**

There are two colleges in the town college A and college B. There are 500 more students in college A than in college B. The ratio of the boys to that of the girls in college A is 17 : 11 and that in college B is 9 : 14. The ratio of the number of Computer Science, Electronics and Mechanical students in college A and college B are 3 : 11 : 14 and 8 : 8 : 7 respectively. The number of Mechanical students in A is twice that in B.

**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

How many students are there in college A?

Answer: 2800

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**Explanation** :

Let the number of students be:

CS E M

College A 3x 11x 14x

College B 8y 8y 7y

The number of commerce students in A is twice that in B

∴ 14x = 2 × 7x

⇒ x = y

∴ Total number of students in A = 28x and in B = 23x

Also, 28x = 23x + 500

⇒ x = 100

Number of students in college A = 28x = 2800

Hence, 2800.

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

How many girls are there in the two colleges together?

Answer: 2500

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**Explanation** :

Consider the solution to first question of this set.

Number of girls in college A = 11/28 × 2800 = 1100

Number of girls in college B = 14/23 × 2300 = 1400

∴ Total number of girls in the two colleges together = 1100 + 1400 = 2500

Hence, 2500.

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

If $\frac{pa}{b+c}=\frac{pb}{c+a}=\frac{pc}{a+b}$ = 𝓁, then what is the value of ‘𝓁’ if p ≠ 0, a + b + c ≠ 0

- (a)
p

- (b)
p/3

- (c)
p/2

- (d)
p/4

Answer: Option C

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**Explanation** :

pa/(b + c) = pb/(c + a) = pc/(a + b) = 𝓁

⇒ pa/(b + c) = 𝓁; pb/(c + a) = 𝓁; pc/(a + b) = 𝓁

So, pa = b𝓁 + c𝓁,

pb = c𝓁+a𝓁,

pc = a𝓁 + b𝓁

Now adding these 3 equations, p(a + b + c) = 𝓁(b + c + c +a + a + b)

⇒ p(a + b + c) = 2𝓁(a + b + c)

As a + b + c ≠ 0,

p = 2𝓁

∴ 𝓁 = p/2

Hence, option (c).

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A bag contains 3 kinds of coins in the following ratios : The number of 5 dollar coins : the number of 2 dollar coin : the number of 1 dollar coins = 3 : 2 : 1. If the total value of the coins is Dollar 240 only, find the number of 1 dollar coins.

- (a)
11

- (b)
12

- (c)
13

- (d)
14

Answer: Option B

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**Explanation** :

Value = No. of coins × denomination in the amount for the type of coin.

Thus, combined ratio of value is 5 × 3 : 2 × 2 : 1 × 1 = 15 : 4 : 1 Then Rs. 1 coin is 1/20 × 240 = Rs. 12 i.e. 12 coins

Hence, option (b).

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

The monthly telephone bill has a fixed tariff for upto 100 outgoing calls. Outgoing call in excess of 100 are charged at a certain fixed rate per call. The monthly bills of Rajesh and Suraj who made 196 outgoing calls and 436 outgoing calls respectively were Rs. 600 and Rs. 900 respectively. Find the monthly bill for a person who has made 320 outgoing calls (in Rs. )

Answer: 755

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**Explanation** :

Let the fixed tariff be F and the number of outgoing calls be C.

Let the charge per call for calls in excess of 100 be k.

If C > 100, the call charge is in the form k(C - 100).

600 = F + k(96) → (1)

900 = F + k(336) → (2)

Subtracting (1) from (2), 300 = 240k

k = 5/4 & F = 480

Required monthly bill = F + k(220) = 480 + (5/4) × 220 = Rs. 755

Hence, 755.

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

P varies directly as the sum of the two quantities Q and R. Q in turn varies directly as x and R varies inversely as x. When x = 4, P = 12 and when x = 8, P = 18. Find the value of P when x = 16.

- (a)
24

- (b)
27

- (c)
32

- (d)
33

Answer: Option D

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**Explanation** :

P α (Q+R)

⇒ P = k(Q+R)

Also,Q α x ⇒ Q =k_{1}x

and, R α 1/x ⇒ R = k_{2}/x

P = kk_{1}x + kk_{2}/x

= p_{1}x + p_{2}/x (where p_{1} and p_{2} are constants)

⇒ 4p_{1 }+ p_{2}/4 = 12

⇒ 16p_{1 }+ p_{2 }= 48 …(1)

and, 8p_{1 }+ p_{2}/8 = 18

⇒ 64p_{1 }+ p_{2 }= 144 …(2)

Solving (1) and (2),

we get∶ p_{1 }= 2 and, p_{2 }= 16

∴ A = 2x + 16/x

When, x = 16, A = 32 + 1 = 33

Hence, option (d).

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**Answer the next 2 questions based on the information given below.**

In CAT exam, there are 3 sections containing 24, 22 and 24 questions respectively. Each question carries 3 marks each. Wrong answer carries 1 negative mark. There is no negative or positive marking for an un attempted question. A student attempted the questions from sections A, B and C in the ratio 8 : 6 : 7. The numbers of correct answers are in the ratio 11 : 9 : 8, respectively. The student got 64 marks in section A.

**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

How many marks will he get in section B?

Answer: 54

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**Explanation** :

Since he got 64 marks in section A, he must have attempted all 24 questions and got 22 correct and 2 wrong.

∴ His marks = 22 × 3 – 2 × 1 = 64.

Ratio of number of questions attempted in three sections = 8 : 6 : 7

Number of questions attempted in section 1 = 24

∴ Number of questions attempted in section 2 = 18

∴ Number of questions attempted in section 3 = 21

Ratio of number of questions correct in three sections = 11 : 9 : 8.

Number of questions correct in section 1 = 22

∴ Number of questions correct in section 2 = 18

∴ Number of questions correct in section 3 = 16

∴ Marks in section B = 18 × 3 = 54

Hence, 54.

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

How many marks will the student get in section C?

Answer: 43

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**Explanation** :

Consider the solution to first question of this set.

In section C, he will get = 16 × 3 – 5 × 1 = 43 marks

Hence, 43.

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**PE 3 - Ratio | Arithmetic - Ratio, Proportion & Variation**

If a, b and c are in continued proportion, then which of the following equals a : c?

- (a)
a

^{2}: b^{2} - (b)
(c

^{2}– b^{2}) : (b^{2}– a^{2}) - (c)
b

^{2}: c^{2} - (d)
More than one of theabove

- (e)
All of (a), (b) and (c)

Answer: Option D

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**Explanation** :

As a, b and c are in continued proportion,

a/b = b/c

⇒ b^{2} = ac

**Option (a)**: a^{2} : b^{2} = a^{2} : ac = a : c

∴ Option (a) is correct.

**Option (b)**: (c^{2} – b^{2}) : (b^{2} – a^{2})

= (c^{2} – ac) : (ac – a^{2})

= c(c - a) : a(c - a)

= a : c

∴ Option (b) is not correct.

**Option (c)**: b^{2} : c^{2} = ac : c^{2} = a : c

∴ Option (c) is correct.

Hence, option (d).

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