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Check NowA box contains 5 apples, 7 oranges and 11 pineapples. How many fruits should one pick from the box to have at least 4 fruits of the same kind?
The maximum number of fruits one can pick without having 4 fruits of the same kind equals 9 when x=y=z=3.
So, if one picks 10 fruits, there will be at least 4 fruits of the same kind.
In a two-digit number, the difference between the digits is 4, and 10 times the number is equal to 14 times the sum of the number obtained by adding the digits of the original and the number formed by reversing the digits. Find the number.
Let the number be x y. Hence $$|x-y| = 4$$ and 10(10x+y)=14(10y+x+x+y)= 14(11y+2x) = 154y+28x => 100x+10y=154y+28x = 72x = 144y => x=2y.
Hence y=4 => x=8. Hence the number is 84
The expression $$(a - b)^3 + (b - c)^3 + (c - a)^3$$ can be factorized as
If X+ Y+Z = 0 , then $$X^3+Y^3+Z^3=3XYZ$$
Let a-b = X, b-c = Y, c-a = Z
We can see that X+Y+Z=0
so $$(a - b)^3 + (b - c)^3 + (c - a)^3$$ = 3(a-b)(b-c)(c-a)
B is the correct answer.
Pune taxi services has a fixed charge plus a variable charge based on the distance covered. For travelling 10km the total fare paid by Somu is Rs.150 and for a journey of 15km the total fare paid by Ramu is RS220. Then Tomy will pay _as the total fare for travelling a distance of 25 km.
Let the fixed charge be Rs. x and variable charge/km is Rs. y.
As per the question,
x + 10y= 150 and x+ 15y = 220
On solving both the equations, we get 5y = 220-150= 70
y = Rs. 14/km and x = Rs. 10
Charge for 25 km = x+ 25y
= 10+25*14 = Rs. 360
C is the correct answer.
Nisha went to buy three types of stationery products, each of them were priced at Rs. 5, Rs, 2 and Rs. 1 respectively. She purchased all three types of products in more than one quantity and gave Rs. 20 to the shopkeeper. Since the shopkeeper had no change with him/her; he/she gave Nisha three more products of price Rs. 1 each. Find out the number of products with Nisha at the end of the transaction.
Let x, y, z be the number of items purchased by Nisha of price Rs. 5, 2, 1 respectively.
5x+2y+z = 20
Since the shopkeeper had no change with him/her; he/she gave Nisha three more products of price Rs. 1 each i.e Nisha purchased items of Rs. 17
5x+2y+z =17
She purchased all three types of products in more than one quantity,
If she purchases three items of Rs. 5, she cannot purchase 2 items of Rs 2 and 1
So Nisha purchases 2 items of Rs. 5.
If Nisha purchases 3 items of Rs 2., then she cannot purchase more than 1 item of Rs. 1.
Hence she must have purchased 2 items of Rs. 2 and 3 items of Rs. 1.
The number of items Nisha will have at the end of the transaction = 2+2+3+3(due to no change)
= 10
D is the correct answer.
Sumit was doing a multiplication, but by mistake instead of taking 25 as one of the multiplier, he took 34. Now, because of this mistake, the answer was 405 more than the correct answer. Find out the answer that Sumit arrived at.
Let the other multiplier be p
Instead of calculating 25*p, he calculated 34*p
It is given that because of the mistake, the answer was 405 more than the correct answer.
34p-25p=405
9p = 405
p=$$\ \frac{\ 405\ }{9}$$
Now we have to find the value of 34p
=34*$$\ \frac{\ 405\ }{9}$$
=1530
A is the correct answer.
David is trying to solve the expression :-
$$\frac{(4)^2 \times 2^{n + 1} - 4 \times 2^n}{(4)^2 \times 2^{n + 2} - 2 \times 2^{n + 2}}$$
And you help him to do the same and finally arrive at the answer with correct to one decimal which would be -
(Note:- DO NOT include spaces in your answer)
$$\frac{(4)^2 \times 2^{n + 1} - 4 \times 2^n}{(4)^2 \times 2^{n + 2} - 2 \times 2^{n + 2}}$$
$$\frac{(2)^4 \times 2^{n + 1} - 2^2 \times 2^n}{(2)^4 \times 2^{n + 2} - 2 \times 2^{n + 2}}$$
$$\frac{(2)^3 \times 2^{n}*2 -2 \times 2^n}{(2)^3 \times 2^{n}*2^2 - 2^{n}*2^2}$$
$$\ \frac{\ 2^3\times\ 2-2}{2^3\cdot2^2-2^2}$$
$$\ \frac{\ 14}{28}$$
=0.5
Ramesh is trying to Simplify the expression
$$(p + q)^3 - (p -q)^3 - 6q(p^2 - q^2)$$ and if $$q = 1$$-
You helped him and the solution arrived was..........
($$p^3+3p^2q+3pq^2+q^3$$) - ($$p^3-3p^2q+3pq^2-q^3$$) - ($$6q(p^2 - q^2)$$)
=$$6p^2q+2q^3-6q(p^2-q^2)$$
= 8$$q^3$$
=8 (Since q = 1)
8 is the correct answer.
John's grandfather was five times older to him 5 years ago. He would be two times of his age after 25 years from now. What is the ratio of John's age to that of his grandfather ?
Let age of john 5 years ago be x years
age of his grandfather 5 years ago will be 5x
25 years from now there ages will be x+30 and 5x+30 respectively
Given 5x+30 =2(x+30) => x=10
So there present ages would be x+5=15 and 5x+5=55
Ratio = 3:11
Two numbers are such that the square of greater number is 504 less than 8 times the square of the other. If the numbers are in the ratio 3 : 4. Find the number.
Let the two numbers be 3x and 4x.
According to the question,
$$\ \ \ \left(4x\right)^2=8\left(3x\right)^2-504$$
=> $$\ \ \ 16x^2=72x^2-504$$
=>$$\ \ \ 56x^2=504$$
=>$$\ \ \ x^2=9$$
Or, x=$$\pm\ 3$$
Since, x cannot be -3, it is 3.
Therefore the numbers are 3x and 4x, which is 9 and 12.
The sum of a number and its reciprocal is thrice the difference of the number and its reciprocal. The number is:
Let the number be a
It is given that sum of a number and its reciprocal is thrice the difference of the number and its reciprocal.
a+$$\ \frac{\ 1}{a}$$ = 3(a-$$\ \frac{\ 1}{a}$$ )
2a = $$\ \frac{\ 4}{a}$$
a = $$\ \frac{\ 2}{a}$$
$$\ \ a^2=2$$
a = $$\pm \surd{2}$$
A is the correct answer.
The difference between the value of a number increased by 25% and the value of the original number decreased by 30% is 22. What is the original number?
Let the number be N.
If the number increases by 25%, it becomes 1.25N, and if it decreases by 30%, then it becomes 0.7N. The difference between them is given as 22,
1.25 N - 0.7 N = 22
=> N = 22/0.55 = 40
A number whose fifth part increased by 4 is equal to its fourth part diminished by 10, is :
Let the number be N.
(N/5) + 4 = (N/4) - 10
=> N = 280
A person buys 18 local tickets for Rs. 110. Each first class ticket costs Rs. 10 and each second class ticket costs Rs. 3. What will another lot of 18 tickets in which the number of first class and second class tickets are interchanged cost?
Consider the number of first-class tickets =a and the second-class ticket =b
10a+3b=110
a+b=18
Solving two equations, we get a=8, b=10
After interchanging the number of tickets, the new cost = 10*10+3*8 =124
In a factory, each day the expected number of accidents is related to the number of overtime hour by a linear equation. Suppose that on one day there were 1000 overtime hours logged and 8 accidents reported and on another day there were 400 overtime hours logged and 5 accidents. What is the expected number of accidents when no overtime hours are logged?
Let the number of overtime hours logged be x.
Let the number of accidents reported be A.
So, since the relationship between A and x is linear, we can write A= ax+c
Given, when x=1000, A=8.
So, 8= 1000a+ c--------------(1)
Again, when x= 400, A=5.
So, 5= 400a+ c ----------------(2)
Subtracting Eqn 2 from 1, we get,
600a= 3
=> a= 1/200.
Now putting this value of a in Eqn 1, we find
8= $$\ \frac{\ 1000}{200}+c$$
=>c= 8-5=3.
So, when x=0,
A= c= 3- option B
A rainy day occurs once in every 10 days. Half of the rainy days produce rainbows. What percent of all the days do not produce rainbow?
Let total number of days=20x
Number of rainy days = 2x
Number of rainbow days = x
Number of non rainbow days= 19x
$$\therefore\ $$ Percentage of non rainbow days = $$\frac{19x}{20x}\ \cdot\ 100\ =95$$
A man purchased 40 fruits: apples and oranges for Rs. 17. Had he purchased as many oranges as apples and as many apples as oranges, he would have paid Rs. 15/-. Find the cost of one pair of an apple and an orange.
Let us assume the number of apples purchased by the man to be x.
Then, the number of oranges will be (40-x).
Let the price of each apple be Rs. y and that of an orange be Rs. z
It is given that, xy+ (40-x)z= 17.
=> xy+40z-xz=17 ---------------(1)
Now, if the number of apples and oranges are interchanged, the apples will be (40-x) and the oranges will be x.
Given, the new total cost will be Rs. 15
So, (40-x)y + xz=15
=>40y-xy +xz= 15 ---------------(2)
Adding eqns 1 and 2, we get,
xy+40z-xz+40y-xy+xz=15+17
=> 40(y+z)=32
=>y+z= $$\ \frac{\ 32}{40}=\ \frac{\ 4}{5}$$
Or, the cost of one apple and one orange, x+y= Rs.$$\frac{\ 4}{5}=\ 80\ paise$$
The following line graph shows the ratio of the wheat production to rice production of a state in India over a period 7 years from 1999 - 2006. Answer the questions based on this data.
In how many years wheat production was more than the rice production?
The number of years in which ratio of wheat production to rice production is greater than 1 will be the years where wheat production was greater than the rice production.
So number of years = 3
The years are 2002-2003, 2003-2004. and 2005-2006
Recently my brother and I played chess form chocolates. Who ever lost the game gave the other a chocolate. After the last game we counted the chocolates. I had 20 more chocolates than I started with, although he won 7 games. There is no draw. How many games did we play?
I have won X number of chess games and my brother has won 7 games.
For every game I win I get 1 chocolate and and I give 1 chocolate if my brother wins. Let A be number of chocolates with me initially.
Then , A +X - 7 = A +20
X= 27
Total games played = Games won by me + Games won by my brother = 27+7 = 34
Anand, Binoy, Chetan and Dharma together have Rs 47 with them. Anand and Binoy together have Rs 27; Chetan and Anand have Rs 25 and Dharma and Anand have Rs 23. How much money does Binoy have?
Let a,b,c and d represent the money with Anand, Binoy, Chetna and Dharma respectively.
It is given that a+b+c+d = 47 ......(I)
a+b = 27 ........(II)
a+c = 25 ........(III)
a+d = 23 ........(IV)
Adding eqn (II),(III) and (IV) and subtracting (I)
(a+b)+(a+c)+(a+d) - (a+b+c+d) = (23+25+27-47)
2a = 28 or a = 14
from eqn (II) b = 13
A picture was bought at a certain sum, which was the price paid for its frame. Had the frame cost Rs 100 less and the picture Rs 75 more the price for the frame would have been only half of that of the picture. What is the price of the frame?
Let the cost of Picture be Rs X. Then the cost of frame is also Rs X.
As per question $$\frac{X+75}{2}$$ = $$2\cdot\left(X-100\right)$$
X + 75 = 2X - 200
X = 275
Ram Singh goes to Pushkar Mela with Rs 10000 to buy exactly 100 animals. He finds that cows are sold at Rs 1000, horses at Rs 300 and chicken at Rs 50. How many chicken should be buy to meet his target of 100 animals?
Let x,y and z be the number of cows, horses and chickens, respectively.
Since he has Rs 10,000.
$$1000x+300y+50z$$ = 10,000 ....(I)
$$ x+y+z$$ = 100 ....(II)
300*eqn(II)- eqn(I) gives
$$250z - 700x = 20,000$$
$$z\ =\ \ \frac{\ 20,000+700x}{250}$$
since $$z$$ is an integer then $$Remainder\left(\frac{\ 20,000+700x}{250}\right)$$ = 0
$$Remainder\left(\frac{\ 20,000+700x}{250}\right)$$ = $$Remainder\left(\frac{200x}{250}\right)$$
$$Remainder\left(\frac{200x}{250}\right)$$ = 0 is possible when x takes values 5 ,10, 15 and so on
When x = 5, then z = 94 and y = 1
When x = 10, then z = 108 ( rejected as the total number of animals has to be 100)
For all the values of x greater than 10, the value of z exceeds 100 and does not satisfy the given conditions.
Thus, he has to buy 94 chickens.
A game of football with 11 players lasts for exactly 90 minutes. There are four substitutes that alternate equally. If each player plays for the same length of time, what is the duration?
Let each alternate player plays for x minutes
Thus 4x = 90
x = 22.5 minutes
A boy was asked to multiply a certain number by 53. He multiplied it by 35 and got
his answer less than the correct one by 1206. Find number to be multiplied.
Let number to be multiplied be $$x$$
Number obtained by the boy = $$35x$$
Actual answer = $$53x$$
Difference obtained = $$53x$$ - $$35x$$ = 1206
$$18x$$ = 1206
$$x$$ = 67
When 75% of a no. is added to 75, the result is the same number. The no. is
Let the number be x
As per question
$$\ \frac{\ 75x}{100}$$ + 75 = x
$$\ \frac{\ 3x}{4}$$ + 75 = x
$$\ \frac{\ x}{4}$$ = 75
x = 4*75 = 300
A family has several children. Each boy in the family has as many sisters as brothers but each girl has twice as many brothers as sisters. How many brothers and sisters are there?
Let there be x boys and y girls in the family.
For each of the boys, number of brother = x-1 and number of sister = y
For each of the girls, number of brother = x and number of sisters = y-1
As per question,
x-1 = y ( equation of brother and sister for boys)
x = 2(y-1) (equation of brother and sister for girls
On solving we get x = 4 and y= 3, Hence option C
SNAP Linear Equations is an important topic in the Quantitative, Data Interpretation & Data Sufficiency section of the SNAP 2026 Exam. Questions from this topic test a candidate's understanding of linear relationships, algebraic equations, and variable-based problem solving. These questions often involve solving single-variable and multi-variable equations, word problems, and applications of linear relationships in real-life scenarios. A strong understanding of linear equations helps candidates solve algebra-based questions quickly and accurately. Practicing topic-wise questions and solving problems from SNAP Previous Papers is an effective way to strengthen concepts and improve accuracy. Additionally, attempting SNAP Mock Tests will help candidates understand the latest exam pattern and the types of questions that can be expected in the actual exam. To gain a better understanding of all the topics covered in the exam, candidates should also refer to the SNAP Syllabus.
Yes, Linear Equations is an important algebra topic in SNAP Quantitative Aptitude. Questions from this area test a candidate's ability to solve equations, analyze relationships between variables, and apply algebraic concepts efficiently.
SNAP does not have a fixed number of questions from Linear Equations. However, candidates can expect direct or concept-based questions involving single-variable equations, simultaneous equations, and algebraic applications.
SNAP may include questions on solving linear equations, simultaneous linear equations, word problems, ratio-based equations, age problems, and applications involving multiple variables.
Start by understanding the basic principles of linear equations and algebraic manipulation. Practice different types of equation-solving techniques, solve questions from SNAP Previous Papers, and regularly attempt SNAP Mock Tests to improve speed and accuracy.
The difficulty level is generally easy to moderate. Most questions are straightforward and can be solved quickly with strong conceptual clarity and regular practice.
Cracku's SNAP Linear Equations Questions are designed according to the latest SNAP exam pattern and difficulty level. They provide topic-wise practice questions, detailed solutions, shortcut methods, and performance analysis to help aspirants strengthen their concepts and improve their scores.
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