Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
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Sujay’s father asked him to get the fruits, milk and butter from the market and gave him the money in the denomination of 1 Rupee, 2 Rupee and 5 Rupee coins. Sujay first goes to the grocery shop to buy fruits. At the grocery shop he gives half of his 5 Rupee coins and in return receives the same number of 1 Rupee coins. Next he goes to a dairy shop to buy milk and butter and gives all 2 Rupee coins and in return gets thirty 5 Rupee coins which increases the number of 5 Rupee coins to 75% more than the original number. If the number of 1 Rupee coins now is 50, then number of 1 Rupee and 5 Rupee coins he had originally were
Solution (d)
Let the number of 1 rupee, 2 rupee and 5 rupee coins be x, y and z respectively
At the grocery store he gives half of his 5 Rupee coins and in return receives the same number of 1 Rupee coins i.e. z now become 0.5z and x now becomes x+0.5z
At the dairy store he gives all 2 Rupee coins and in return gets thirty 5 Rupee coins which increases the number of 5
Rupee coins to 75% more than the original number i.e. now y = 0 and 0.5z = 0.5z+30
Also given 0.5z+30 = 1.75z i.e. z = 24
Also given that number of 1 Rupee coins now is 50 i.e. x+0.5z = 50 i.e. x = 38
Therefore the number of 1 Rupee and 5 Rupee coins originally were 38 and 24 respectively.
Solution (d)
Let the number of 1 rupee, 2 rupee and 5 rupee coins be x, y and z respectively
At the grocery store he gives half of his 5 Rupee coins and in return receives the same number of 1 Rupee coins i.e. z now become 0.5z and x now becomes x+0.5z
At the dairy store he gives all 2 Rupee coins and in return gets thirty 5 Rupee coins which increases the number of 5
Rupee coins to 75% more than the original number i.e. now y = 0 and 0.5z = 0.5z+30
Also given 0.5z+30 = 1.75z i.e. z = 24
Also given that number of 1 Rupee coins now is 50 i.e. x+0.5z = 50 i.e. x = 38
Therefore the number of 1 Rupee and 5 Rupee coins originally were 38 and 24 respectively.
Read the following information carefully to answer the questions given below it.
Mr Ram’s family is a traditional joint family having six persons from three generations. Each member of the family has different food preference and they support different sports / games. Only two couples are there in the family. Rajesh likes continental food and his wife neither likes dry fruits nor supports gymnastics. The person who likes egg supports Rugby and his wife likes traditional food. Mona is mother-in-law of Sona and she supports Athletics. Varad is grandfather of Taman and Taman, who likes Punjabi food, supports Basketball. Nanditha is granddaughter of Mona and she supports Badminton. Nanditha’s mother supports horse riding.
Identify the correct pair of two couples from the following
Solution (a)
We are given that,
Varad is grandfather of Taman. Nanditha is grand-daughter of Mona
Thus, Taman and Nanditha belongs to the 3rd generation whereas Mona and Varad belong to the 1st generation.
Thus, Rajesh and Sona belong to the 2nd generation.
Mona is the mother-in-law of Sona and thus, the family tree is as shown
Now, from the given information we can say that
Mona – Traditional Food, Athletics
Varad – Egg, Rugby
Sona – Horse riding
Rajesh – Continental Food
Taman – Punjabi Food, Basketball
Nanditha – Badminton
Thus, Mona – Varad and Rajesh – Sona are the 2 couples.
Hence, option a is correct
Solution (a)
We are given that,
Varad is grandfather of Taman. Nanditha is grand-daughter of Mona
Thus, Taman and Nanditha belongs to the 3rd generation whereas Mona and Varad belong to the 1st generation.
Thus, Rajesh and Sona belong to the 2nd generation.
Mona is the mother-in-law of Sona and thus, the family tree is as shown
Now, from the given information we can say that
Mona – Traditional Food, Athletics
Varad – Egg, Rugby
Sona – Horse riding
Rajesh – Continental Food
Taman – Punjabi Food, Basketball
Nanditha – Badminton
Thus, Mona – Varad and Rajesh – Sona are the 2 couples.
Hence, option a is correct
In each of the following letter series, some of the letters are missing, which are given below it. Choose the correct alternative.
D_F_DEE_D_EF_DE_F
Solution (c)
DEF/FDE/EFD/DEF/FDE/EF
Here, DEF are rotating in cyclic order.
Hence, option c is correct
Solution (c)
DEF/FDE/EFD/DEF/FDE/EF
Here, DEF are rotating in cyclic order.
Hence, option c is correct
In each of the following letter series, some of the letters are missing, which are given below it. Choose the correct alternative.
_OPO_QOPQ_RQPO_POR_O
Solution (d)
By using all the options in the given series
Option a: A O P O P Q O P Q R R Q P O Q P O R O O
Option b: Q O P O P Q O P Q O R Q P O R P O R O O
Option c: Q O P O P Q O P Q R R Q P O O P O R O O
No logical pattern can be seen in any of the option a, b or c.
Hence, option d is correct.
Solution (d)
By using all the options in the given series
Option a: A O P O P Q O P Q R R Q P O Q P O R O O
Option b: Q O P O P Q O P Q O R Q P O R P O R O O
Option c: Q O P O P Q O P Q R R Q P O O P O R O O
No logical pattern can be seen in any of the option a, b or c.
Hence, option d is correct.
Raj bought a certain number of pastries flavoured pineapple, mango and black-forest from the bakery for his birthday party, giving for each pastry as many rupees as there were pastry of that kind; altogether he bought 23 pastries and spent Rs 211. Find the number of each kind of pastry that he bought, if mango pastries are cheaper than pineapple pastries and dearer than black-forest pastries.
Solution (b)
Given that, the cost of a type of pastry is equal to the quantity bought
If x, y, z are the number of pastries bought of each type, then x+y+z = 23 and their cost = 211 Rs
Using options,
Option a = 10, 9, 4 i.e. 100+81+16 = 197
Option b= 11, 9, 3 i.e. 121+81+9 = 211
Option c = 10, 8, 5 i.e. 100+64+25 = 189
Option d = 11, 8, 4 i.e. 121+64+16 = 201
Therefore option b is correct
Solution (b)
Given that, the cost of a type of pastry is equal to the quantity bought
If x, y, z are the number of pastries bought of each type, then x+y+z = 23 and their cost = 211 Rs
Using options,
Option a = 10, 9, 4 i.e. 100+81+16 = 197
Option b= 11, 9, 3 i.e. 121+81+9 = 211
Option c = 10, 8, 5 i.e. 100+64+25 = 189
Option d = 11, 8, 4 i.e. 121+64+16 = 201
Therefore option b is correct
