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Arithmetic I

Ratio & Partnership

ALL MODULES

CAT 2025 Lesson : Ratio & Partnership - Past Questions

8. Past Questions

Question 1

A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of

6 : 7 : 8 : 9 : 10

. In all papers together, the candidate obtained

60 \%

of the total marks. Then the number of papers in which he got more than

50 \%

marks is
[CAT 2001]

		$2$
		$3$
		$4$
		$5$

Observation/Strategy

1

) As all the values are in relative terms, i.e., in ratio or percentage, we can assume a value for total marks.

2

) As

6, 7, 8, 9

and

10

are in AP, the middle term is the average. The student must have scored

60 \%

in paper with weight

8

. So, the student definitely scored

60 \%

or more in

3

of the papers (those with weights

8, 9

and

10

).

Let the total marks for each paper be

100

. Total marks across

5

papers

= 500.

Marks scored by the student in the

5

papers

= 60 \%

500 = 300

Marks scored in

1^{\text{st}}

paper

\dfrac{6}{40} \times 300 = 45

Marks scored in

2^{\text{nd}}

paper

= \dfrac{7}{40} \times 300 = 52.5

∴ The student scored more than

50

marks in 4 papers (

2^{\text{nd}}, 3^{\text{rd}}, 4^{\text{th}}

and

5^{\text{th}}

papers)

Alternatively

Marks scored in the

3^{\text{rd}}

paper with weight

8

60

.

In the

1^{\text{St}}

and

2^{\text{nd}}

papers, the weights are

6

and

7

respectively. These weights are

\dfrac{2}{8}

and

\dfrac{1}{8}

less than

8

.

∴ Marks scored will also be

\dfrac{2}{8}

and

\dfrac{1}{8}

less than

60

\dfrac{1}{8} \times 60 = 7.5

Marks in

2^{\text{nd}}

subject

= 60 - 7.5 = 52.5

Marks in

1^{\text{st}}

subject

= 60 - 2 \times 7.5 = 45

∴ The Student scored more than

50

in 4 papers.

Answer:

(3)

Question 2

Ravindra and Rekha got married 10 years ago, their ages were in the ratio of

5 : 4

. Today Ravindra’s age is one sixth more than Rekha’s age. After marriage, they had

6

children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of

3 : 2 : 1

. What is the largest possible value of the present total age of the family?
[IIFT 2014]

		$79$
		$93$
		$101$
		$107$

Observation/Strategy

1

) The only absolute value given is the number of years for which they've been married, which is

10

years. The ages of the kids should be less than

10

2

) The ratio of ages of oldest and youngest kids is

3 : 1

. So, the oldest kid's age is a multiple of

3

. The largest multiple of

3

less than

10

is 9 years.

The ages of the triplets, twins and sixth child should be

9

years,

6

years and

3

years.
Sum of ages of kids

= 3 \times 9 + 2 \times 6 + 3 = \bm{42}

As Ravindra's present age is one-sixth more than that of Rekha, let their present ages be

7x

and

6x

respectively.

10

years back ⇒

\dfrac{7x - 10}{6x - 10} = \dfrac{5}{4}

⇒

28x - 40 = 30x - 50

⇒

x = 5

Sum of Ravindra and Rekha's ages

= 7x + 6x = 13x = 13 \times 5 = \bm{65}

Total age of family

= 42 + 65 = \bm{107}

Answer:

(4)

107

Question 3

Ram and Shyam form a partnership (with Shyam as working partner) and start a business by investing Rs.

4,000

and Rs.

6,000

respectively. The conditions of partnership were as follows:
- In case of profits till Rs.

200,000

per annum, profits would be shared in the ratio of the invested capital.
- Profits from Rs.

200,001

till Rs.

400,000

Shyam would take

20 \%

out of the profit, before the division of remaining profits, which will then be based on ratio of invested capital.
- Profits in excess of Rs.

400,000

, Shyam would take

35 \%

out of the profits beyond Rs.

400,000

, before the division of remaining profits, which will then be based on ratio of invested capital.

If Shyam’s share in a particular year was Rs.

367,000

, which option indicates the total business profit (in Rs.) for that year?
[XAT 2012]

		$520,000$
		$530,000$
		$540,000$
		$550,000$
		None of the above

Observation/Strategy

1

) Ratio of capital invested

= 4,000 : 6,000 = 2 : 3

2

) We start populating Shyam's share for each profit bracket and find when it totals to Rs.

367,000.

Share of Shyam in the
- first Rs.

2

lakhs of profit

= \dfrac{3}{5} \times 200,000 = 120,000

- next Rs.

2

lakhs of profit

= 40,000 + \dfrac{3}{5} \times 160, 000 = 136,000

Share of Shyam in first Rs.

4

lakhs

= 120,000 + 136,000 =

Rs.

256,000

Share of Shyam in profit over Rs.

4

lakhs

= 367,000 - 256,000 =

Rs.

111,000

Let the profits over Rs. 400,000 be

\bm{x}

.

∴ Shyam's share in profit over Rs.

4

lakhs

= 0.35x + \dfrac{3}{5} \times 0.65x = 111,000

⇒

0.74x = 111,000

⇒

x = 150,000

∴ Total profit

= 400,000 + 150,000 =

Rs. 550,000

Answer:

(4)

550,000

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