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% of 500=300
Marks scored in 1st paper 406×300=45
Marks scored in 2nd paper =407×300=52.5
∴ The student scored more than 50 marks in 4 papers (2nd,3rd,4th and 5th papers)
Alternatively
Marks scored in the 3rd paper with weight 8 is 60.
In the 1St and 2nd papers, the weights are 6 and 7 respectively. These weights are 82 and 81 less than 8.
∴ Marks scored will also be 82 and 81 less than 60.
81×60=7.5
Marks in 2nd subject =60−7.5=52.5
Marks in 1st subject =60−2×7.5=45
∴ The Student scored more than 50 in 4 papers.
Answer: (3) 4
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 =3×9+2×6+3=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 ⇒ 6x−107x−10=45
⇒ 28x−40=30x−50
⇒ x=5
Sum of Ravindra and Rekha's ages =7x+6x=13x=13×5=65
Total age of family =42+65=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 =53×200,000=120,000
- next Rs. 2 lakhs of profit =40,000+53×160,000=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 x.
∴ Shyam's share in profit over Rs. 4 lakhs =0.35x+53×0.65x=111,000
⇒ 0.74x=111,000