Expected CLAT 2020 Quantitative Technique Syllabus
Number System + HCF & LCM
Sample Question: A six digit number 123A45 is divisible by 9.Find the value of A.
Solution: For divisibility by 9, the sum of digits of the number has to be divisible by 9, therefore 1+2+3+A+4+5=15+A. For this sum (15+A) to be divisible by 9 the number A has to be 3 as it will only suffice the condition.
Sample Question: the average of 5 numbers a, a+2, a+4, a+6 and a+8 is 11. Find the average of last 3 numbers.
Solution: Sum = a+a+2+a+4+a+6+a+8 = 5a+20
Average = 5a+20 / 5 = a+4 = 11
Therefore, a = 7
Mean of last three numbers = middle number in these thee, i.e., a+6 = 7+6 = 13
Mixtures and Allegations
Sample Question: In what ratio Two kinds of Rice at Rs. 115 and Rs. 124 per kg be mixed so that selling at Rs. 120 you make a profit of 25 per cent.
Solution: Rs. Rs.
The ratio would be 4:5
Ratio and Proportion
Sample Question: Ratio of the salary of A and B is 4:5. A’s salary is increased by 10 per cent and B’s salary is decreased by 20 per cent. Find the new ratio.
Solution: let salaries of A and B be 4000 and 5000. If A’s salary increases by 10 per cent then his new salary would be 4000*1.1 = 4400, and B’s salary decreases by 20 per cent then his new salary would be 5000*0.8 = 4000.
New ratio = 4400/4000 = 11/10
Sample Question: 65 % of a number is 21 less than 4/5 of that number. Find the number.
Solution: Let the number be N,
Then, (4/5) N – (65% * N) = 21 =>( 4/5 – 13/20) N = 21 => ( 3/20) N = 21 => N =140
Simple Interest & Compound Interest
Sample Question: A sum of money is put in CI for 2 years. If the interest paid is half-yearly then it would have fetched Rs.482 more than earlier. Find the sum.
Solution: Let the Principal be Rs. 1000
Annual compound = A = 1000[1+20/100]2
Half-yearly compound = A1 = 1000[1+10/100]4
Difference between the two = 1440 – 1464.1
Positive difference = 24.1
The difference is 24.1 for principal 1000 therefore for 482 the principal would be Rs. 20000
Profit & Loss
Sample Question: A notebook is purchased at Rs. 20 and sold for Rs. 28. What is the profit per cent on the notebook?
Solution: Profit = SP-CP = 28-20 = 8
Profit Per cent = (8 / 20) *100 = 40 per cent.
Time Speed and Distance
Sample Question: Ram is waking at a speed of 15km/hr and after every hour he takes a break of 3 mins. In how much time he will cover 4 km.
Solution: In 1 hr he walks 15 km than to travel a distance of 4 km he would take (s=d*t) 4 min. therefore, time taken to cover 4 km including rest time would be 4+3+4+3+4+3+4+3=28 min.
Time and Work
Sample Question: P can do a work in 20 days, Q can do the same work in 24 days. With the help of R, they finished the work in 8 days. How many days would R take to complete the work alone?
Solution: When two or more people are completing the work together the time taken by the individual person is added inversely and equated to the inverse of time taken by all the people together. i.e. (1/P)+(1/Q)+(1/R) = 1/8. By solving the above equation you get time taken by R to complete the task alone is 30 days
Linear Equations in two variables
Sample Question: Difference between the age of A and B is 16 years. In 6 years the elder one will be thrice as old as the younger one. Find their present ages.
Solution: Let ages of A and B be a& b Respectively.
a-b = 16 ; Suppose a is elder, a= 3b ; 3b-b = 16 => b=8 and a= 24
Sample Question: Sum of squares of two consecutive even integers is 52. Find those numbers.
Solution: Two even consecutive integer = x and x+2
X2 + (X+2)2 = 72 => x2 + 2x + 2 = 26 => x2 + 2x-24 = 0
Solving the quadratic equation, value of x =4
Numbers = 4 and 6
Square roots and Cute Roots
Sample Question: √(1 + x/144) = 13/12. Find x
Solution: (1 + x/144) = (13/12)2
x/144 = (169-144)/144 = 25/144
Sample question: find the value of the given equation: log2 2+ log2 22 + log2 23 …… +log2 2n
Solution: log2 2+ log2 22 + log2 23 …… +log2 2n
log2 2 + 2log2 2 + 3 log2 2 ……+ n log2 2
(log 2 with base 2 is 1)
1+2+3+….n = n(n+1) / 2
Lines & Angles(2 dimension)
Sample Question: In a Δ ABC, ½ ∠A + 1/3 ∠C + ½ ∠B = 80°. Find ∠C
Solution: 3 ∠A+ 2∠C + 3∠ B = 480
3(∠A+∠B) + 2∠C = 480 … i
Also,[ ∠A+ ∠B + ∠C = 180 ] * 3
3∠A+ 3∠B + 3 ∠C = 540 … ii
Eq ii – i, ∠C = 60
Sample Question: For an isosceles triangle, the perimeter is 30 cm. Find the area.
Solution: Let length sides of the triangle be a, a and b ( isosceles triangle)
a+a+b = 30 => 2a + b = 30 also b2 = 2 a2
b = √2. a =>putiing in other eq : 2a + √2. a = 30 => a(2+√2) = 30
a=8.78 and b = 12.42
Solid & Mensuration
Sample Question: Area of rectangle is 4 times that of a square. Length of the rectangle is 80 cm and breadth is 3 times the size of the square. Find the side of the square
Solution: Let a be side of square, l be length of rectangle and b be the breadth of the rectangle.
l * b = 4 a2 => 80 * 3 a = 4 a2 => a =60 cm
Trigonometry (very basic concepts) and Height and Distance
Sample Question: A person standing at the bank of a river observes the angle of elevation of the top of the tree on the opposite side is 45°. After moving 20 m away from the bank the angle of elevation changes to 30°. Find the height of the tree.
A x B 20 C
AD= h = heights of the tree, that person was earlier standing at B and then moved to C
Tan 45 =1 = h/x => h=x
Tan 30 = 1/√3 = h/ x+20 => 1/√3= h/h+20
Therefore, h= 10(√3+1)m
Modern Mathematics (Low Priority)
Sequence & Series (AP)
Sample Question:In an AP, Ratio of 2nd and th term is 1/3. 5th term is 11, find 15th term.
Solution: a+d / a+6d = 1/3 => 3a +3d = a + 6d => 2a = 3d
5th term, a+4d = 11 => 3d/2 + 4d = 11 => (3+8)/d = 22
d= 2 and a=3 , therefore 15th term, a+14d = 31
Sample Question: A coin tossed for 5 times what is the probability of (HTHHT) this sequence being the toss result. H being head and T being Tail.
Solution: Coin tossed for 5 times, all the possible results are 25
But the sequence can come only once, therefore, probability of receiving this (HTHHT) sequence be 1/25