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Arithmetic I

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Ratio & Partnership

Ratio And Partnership

MODULES

Basics of Ratio & Shortcuts
Properties of Ratio
Comparing Ratios
Combining, Using Variables & Multiplying
Partnership
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Ratio and Proportion 1
-/10
Ratio and Proportion 2
-/10
Ratio and Proportion 3
-/10

PRACTICE

Ratio & Partnership : Level 1
Ratio & Partnership : Level 2
Ratio & Partnership : Level 3
ALL MODULES

CAT 2025 Lesson : Ratio & Partnership - Past Questions

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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 6:7:8:9:106 : 7 : 8 : 9 : 106:7:8:9:10. In all papers together, the candidate obtained 60%60 \%60% of the total marks. Then the number of papers in which he got more than 50%50 \%50% marks is
[CAT 2001]

222
333
444
555

Observation/Strategy
111) As all the values are in relative terms, i.e., in ratio or percentage, we can assume a value for total marks.
222) As 6,7,8,96, 7, 8, 96,7,8,9 and 101010 are in AP, the middle term is the average. The student must have scored 60%60 \%60% in paper with weight 888. So, the student definitely scored 60%60 \%60% or more in 333 of the papers (those with weights 8,98, 98,9 and 101010).

Let the total marks for each paper be
100100100. Total marks across 555 papers =500.= 500.=500.

Marks scored by the student in the
555 papers =60%= 60 \%=60% of 500=300500 = 300500=300

Marks scored in
1st1^{\text{st}}1st paper 640×300=45\dfrac{6}{40} \times 300 = 45406​×300=45

Marks scored in
2nd2^{\text{nd}}2nd paper =740×300=52.5= \dfrac{7}{40} \times 300 = 52.5=407​×300=52.5

∴ The student scored more than
505050 marks in 4 papers (2nd,3rd,4th2^{\text{nd}}, 3^{\text{rd}}, 4^{\text{th}}2nd,3rd,4th and 5th5^{\text{th}}5th papers)

Alternatively

Marks scored in the
3rd3^{\text{rd}}3rd paper with weight 888 is 606060.

In the
1St1^{\text{St}}1St and 2nd2^{\text{nd}}2nd papers, the weights are 666 and 777 respectively. These weights are 28\dfrac{2}{8}82​ and 18\dfrac{1}{8}81​ less than 888.

∴ Marks scored will also be
28\dfrac{2}{8}82​ and 18\dfrac{1}{8}81​ less than 606060.

18×60=7.5\dfrac{1}{8} \times 60 = 7.581​×60=7.5

Marks in
2nd2^{\text{nd}}2nd subject =60−7.5=52.5= 60 - 7.5 = 52.5=60−7.5=52.5
Marks in
1st1^{\text{st}}1st subject =60−2×7.5=45= 60 - 2 \times 7.5 = 45=60−2×7.5=45

∴ The Student scored more than
505050 in 4 papers.

Answer:
(3)(3)(3) 4

Question 2

Ravindra and Rekha got married 10 years ago, their ages were in the ratio of 5:45 : 45:4. Today Ravindra’s age is one sixth more than Rekha’s age. After marriage, they had 666 children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of 3:2:13 : 2 : 13:2:1. What is the largest possible value of the present total age of the family?
[IIFT 2014]

797979
939393
101101101
107107107

Observation/Strategy
111) The only absolute value given is the number of years for which they've been married, which is 101010 years. The ages of the kids should be less than 101010.
222) The ratio of ages of oldest and youngest kids is 3:13 : 13:1. So, the oldest kid's age is a multiple of 333. The largest multiple of 333 less than 101010 is 9 years.

The ages of the triplets, twins and sixth child should be
999 years, 666 years and 333 years.
Sum of ages of kids
=3×9+2×6+3=42= 3 \times 9 + 2 \times 6 + 3 = \bm{42}=3×9+2×6+3=42

As Ravindra's present age is one-sixth more than that of Rekha, let their present ages be
7x7x7x and 6x6x6x respectively.

101010 years back ⇒ 7x−106x−10=54\dfrac{7x - 10}{6x - 10} = \dfrac{5}{4}6x−107x−10​=45​

⇒
28x−40=30x−5028x - 40 = 30x - 5028x−40=30x−50

⇒
x=5x = 5x=5

Sum of Ravindra and Rekha's ages
=7x+6x=13x=13×5=65= 7x + 6x = 13x = 13 \times 5 = \bm{65}=7x+6x=13x=13×5=65

Total age of family
=42+65=107= 42 + 65 = \bm{107}=42+65=107

Answer:
(4)(4)(4) 107

Question 3

Ram and Shyam form a partnership (with Shyam as working partner) and start a business by investing Rs. 4,0004,0004,000 and Rs. 6,0006,0006,000 respectively. The conditions of partnership were as follows:
- In case of profits till Rs.
200,000200,000200,000 per annum, profits would be shared in the ratio of the invested capital.
- Profits from Rs.
200,001200,001200,001 till Rs. 400,000400,000400,000 Shyam would take 20%20 \%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.
400,000400,000400,000, Shyam would take 35%35 \%35% out of the profits beyond Rs. 400,000400,000400,000, 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.
367,000367,000367,000, which option indicates the total business profit (in Rs.) for that year?
[XAT 2012]

520,000520,000520,000
530,000530,000530,000
540,000540,000540,000
550,000550,000550,000
None of the above

Observation/Strategy
111) Ratio of capital invested =4,000:6,000=2:3= 4,000 : 6,000 = 2 : 3=4,000:6,000=2:3
222) We start populating Shyam's share for each profit bracket and find when it totals to Rs. 367,000.367,000.367,000.

Share of Shyam in the
- first Rs.
222 lakhs of profit =35×200,000=120,000= \dfrac{3}{5} \times 200,000 = 120,000=53​×200,000=120,000

- next Rs.
222 lakhs of profit =40,000+35×160,000=136,000= 40,000 + \dfrac{3}{5} \times 160, 000 = 136,000=40,000+53​×160,000=136,000

Share of Shyam in first Rs.
444 lakhs =120,000+136,000== 120,000 + 136,000 ==120,000+136,000= Rs. 256,000256,000256,000

Share of Shyam in profit over Rs.
444 lakhs =367,000−256,000== 367,000 - 256,000 ==367,000−256,000= Rs. 111,000111,000111,000

Let the profits over Rs. 400,000 be
x\bm{x}x.

∴ Shyam's share in profit over Rs.
444 lakhs =0.35x+35×0.65x=111,000= 0.35x + \dfrac{3}{5} \times 0.65x = 111,000=0.35x+53​×0.65x=111,000
⇒
0.74x=111,0000.74x = 111,0000.74x=111,000

⇒
x=150,000x = 150,000x=150,000

∴ Total profit
=400,000+150,000== 400,000 + 150,000 ==400,000+150,000= Rs. 550,000

Answer:
(4)(4)(4) 550,000

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