कैट 2024 स्लॉट 3 प्रश्न (मेमोरी आधारित)
Q 1. 3a=4,4b=5,5c=6,6d=7,7e=8,8f=9. abcdef ⇒?
Solution:
From 3a=4, take log:
a=log34
From 4b=5 :
b=log45
From 5c=6 :
c=log56
From 6d=7 :
d=log67
From 7e=8 :
e=log78
From 8f=9 :
f=log89
abcdef=(log34)(log45)(log56)(log67)(log78)(log89)
Using the change of base formula logab⋅logbc=logac, the product simplifies:
(log34)(log45)(log56)(log67)(log78)(log89)=log39
log39=log3(32)=2
Q 2. Upto 500, how many numbers has distinct digits
Solution:
Case 1: 1-digit numbers
There are 1 -digit numbers from 1 to 9 (excluding 0 since 0 is not counted in the distinct digits range here): 9 numbers.
Case 2: 2-digit numbers
For a 2-digit number AB :
- The first digit A (tens place) has 9 choices (1 to 9 ).
- The second digit B (units place) must be different from A, so it has 9 choices ( 0 to 9, excluding A ).
Thus, 9×9=81 numbers
Case 3: 3-digit numbers less than 500
For a 3-digit number ABC, with A as the hundreds place:
- A can be 1,2,3, or 4 (since the number must be less than 500 ). This gives 4 choices for A.
- B (tens place) must be different from A, so it has 9 choices ( 0 to 9, excluding A ).
- C (units place) must be different from both A and B, so it has 8 choices.
Thus, 4×9×8=288 numbers
So, the total count of numbers up to 500 with distinct digits is 9+81+288=378
Q 3. Avg of 3 numbers, is 28 when the largest is decreased by 10, smallest is increased by 7, Avg becomes 2 more than middle term Keeping the order unchanged, the Difference between largest and smallest number is now 64. Find the original largest number.
Solution:
Let a be the smallest, b be the middle and c be the largest number.
Average of 3 numbers is 28
a+b+c3=28a+b+c=84
When the largest number is decreased by 10 and the smallest number is increased by 7, the new average becomes 2 more than the middle term:
- The new numbers are a+7,b, and c−10.
- The new average is:
(a+7)+b+(c−10)3=b+2
Simplify the left-hand side:
a+b+c−33=b+2
Substitute a+b+c=84 :
84−33=b+2813=b+227=b+2b=25
The difference between the largest and smallest number after modifications is 64:
- The modified numbers are a+7 and c−10.
- The difference between them is:
(c−10)−(a+7)=64
Simplify:
c−a−17=64c−a=81
Solve for a and c
Substituting b=25 in a+b+c=84 :
a+25+c=84a+c=59
We also have c−a=81.
Let's solve these two equations:
1. a+c=59
2. c−a=81
Add the two equations:
(a+c)+(c−a)=59+812c=140c=70
Now substitute c=70 in a+c=59 :
a+70=59a=−11
The original largest number is c=70
Q 4. Gopi mark up the price by 20%, Ram gets 10% discount thus saving Rs 15. Find Profit mode by Gopi?
Solution:
Let the cost price (C.P.) of the item be x.
Marked Price (M.P.)
Gopi marks up the price by 20% :
M.P. =x+0.2x=1.2x
Ram gets a 10% discount on the marked price:
S.P. =1.2x−0.1(1.2x)=1.2x−0.12x=1.08x
The discount given to Ram is:
Discount = M.P. − S.P. =1.2x−1.08x=0.12x
From the problem, Ram saves Rs 15 as the discount:
0.12x=15⟹x=150.12=125
The cost price is x=125, and the selling price is:
S.P. =1.08x=1.08×125=135
Profit is the difference between the selling price and the cost price:
Profit = S.P. − C.P. =135−125=10
Q 5. Sum of all distinct values of 10x+410x=812
Solution:
Let y=10x. Then, the equation becomes:
y+4y=812
y2+4=812y
2y2−81y+8=0
Use the quadratic formula:
y=−b±b2−4ac2a,a=2,b=−81,c=8.y=81±812−4(2)(8)2(2).y=81±6561−644.y=81±64974.
Thus, the solutions for y are:
y1=81+64974,y2=81−64974.
Since y=10x, the corresponding solutions for x are:
x1=log10(81+64974),x2=log10(81−64974)
Using the property of logarithms:
log10(a)+log10(b)=log10(a⋅b)
Here, a=81+64974,b=81−64974
The product a⋅b is:
a⋅b=(81+6497)(81−6497)16a⋅b=812−(6497)216a⋅b=6561−649716a⋅b=6416=4
Thus, log10(a)+log10(b)=log10(4)
log10(4)=log10(22)=2log10(2).
Q 6. |x+y|+|x−y|=2 No. of pairs of (x,y) possible?
Solution: The absolute values |x+y| and |x−y| depend on the signs of x+y and x−y. We consider the following cases:
Case 1: x+y≥0 and x−y≥0 :
|x+y|=x+y,|x−y|=x−y
Substituting into the equation:
(x+y)+(x−y)=2⟹2x=2⟹x=1
Substituting x=1 into x+y≥0 and x−y≥0 :
1+y≥0⟹y≥−1,1−y≥0⟹y≤1
So, y∈[−1,1].
This gives 3 integer pairs: (1,−1),(1,0),(1,1).
Case 2: x+y≥0 and x−y≤0 :
|x+y|=x+y,|x−y|=−(x−y)
Substituting into the equation:
(x+y)−(x−y)=2⟹2y=2⟹y=1
Substituting y=1 into x+y≥0 and x−y≤0 :
x+1≥0⟹x≥−1,x−1≤0⟹x≤1
So, x∈[−1,1].
This gives 3 integer pairs: (−1,1),(0,1),(1,1).
Case 3: x+y≤0 and x−y≥0 :
|x+y|=−(x+y),|x−y|=x−y
Substituting into the equation:
−(x+y)+(x−y)=2⟹−2y=2⟹y=−1
Substituting y=−1 into x+y≤0 and x−y≥0 :
x−1≥0⟹x≥1,x+(−1)≤0⟹x≤1
So, x=1.
This gives 1 integer pair: (1,−1).
Case 4: x+y≤0 and x−y≤0 :
|x+y|=−(x+y),|x−y|=−(x−y)
Substituting into the equation:
−(x+y)−(x−y)=2⟹−2x=2⟹x=−1
Substituting x=−1 into x+y≤0 and x−y≤0 :
−1+y≤0⟹y≤1,−1−y≤0⟹y≥−1
So, y∈[−1,1].
This gives 3 integer pairs: (−1,−1),(−1,0),(−1,1).
However, some pairs are repeated:
- (1,−1) appears in Cases 1 and 3.
- (−1,1) appears in Cases 2 and 4.
Thus, there are:
3+3+1+3−2=8 distinct pairs
कैट 2024 स्लॉट 1 प्रश्न (स्मृति आधारित) (CAT 2024 Slot 1 Questions (Memory Based)
CAT 2024 परीक्षा का दोपहर का स्लॉट समाप्त होते ही यह अनुभाग अपडेट कर दिया जाएगा।
CAT 2024 Slot 1 QUANT Questions
1. Find the remainder of 1001007
2. There are 187 fruits in a basket. The ratio of apples to mangoes is in the ratio 5:2. After selling 75 apples and 26 mangoes and half the number of oranges, the leftover apples and oranges was in the ratio 3:2.
Ans: 66
3. If an represents the greatest integer function of n, then sum of a1,a2,a3,…a49,a50 is?
Ans: 217
4. S.P of product is fixed to ensure 40% profit. If product cost 40% less, and have been sold for Rs. 5 less, Then Profit would have been 50%, then the original S.P is:
A) 5
B) 10
C) 20
D) 14
Ans: 14
5. A vessel is full of milk 2/3rd of the milk is removed and replaced with water. This process is repeated 3 more times. Then find ratio of milk and water in the final solution.
6. AB=56,CD=45. Inradius of circle in ADE..
Ans: 10
7. Q. (1/8)k×(1/85)1/3=(1/8)x (1/85)1/k. Find the sum of all posslble values of ' k '
8. For any natural Number ' n ', let an be the largest number not exceeding n, then a1+a2+ a3... +a50=
9. x is a +ve real no,4lox10x+4log100x+ 8log1000x=13, then the greatest integer not exceeding ' x '
10. (1/8)k×(1/85)1/3=(1/8)× (1/85)1/k. Find the sum of all possible values of ' k '
11. Let x,y,z be real numbers and 4(x3+y2+z2)=a,4(x−y−z)=3+a, then find the value of ' a '
