Linear Equations<\/b>\u00a0is a key topic in the CAT<\/strong> Quants <\/strong>section. M<\/span>ake sure you are aware of all the Important Concepts in CAT Quant (Linear Equation)<\/span><\/strong>.<\/span> One must not miss out on the questions on Linear Equations in the QA section. Linear Equations fall under the category of Algebra<\/strong> in the CAT Quants; <\/span>You can also check out these CAT Linear Equation questions<\/span> from the CAT Previous year papers<\/strong><\/span>.<\/span> This post will look into some important Linear Equation questions for CAT. These are good sources of practice for CAT 2022 preparation. If you want to practice these questions, you can download these Important Linear Equation Questions for CAT<\/strong> (with detailed answers) PDF<\/strong> and the video solutions below, which are completely Free.<\/p>\n Download Linear Equations Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?<\/p>\n 1)\u00a0Answer:\u00a062<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the initial number of chocolates be 64x.<\/p>\n First child gets 32x+1 and 32x-1 are left.<\/p>\n 2nd child gets 16x+1\/2 and 16x-3\/2 are left<\/p>\n 3rd child gets 8x+1\/4 and 8x-7\/4 are left<\/p>\n 4th child gets 4x+1\/8 and 4x-15\/8 are left<\/p>\n 5th child gets 2x+1\/16 and 2x-31\/16 are left.<\/p>\n Given, 2x-31\/16=0=> 2x=31\/16 => x=31\/32.<\/p>\n .’. Initially the Gentleman has 64x i.e. 64*31\/32 =62 chocolates.<\/p>\n Question 2:\u00a0<\/b>Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is<\/p>\n a)\u00a05<\/p>\n b)\u00a04<\/p>\n c)\u00a06<\/p>\n d)\u00a07<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given<\/p>\n A+(B+C)\/2=5 => 2A+B+C=10….(i)<\/p>\n (A+C)\/2 +B=7 => A+2B+C=14…..(ii)<\/p>\n (i)-(ii)=> B-A=4 => B=4+A.<\/p>\n Given, A, B, C are positive integers<\/p>\n If A=1=>B=5 => C=3<\/p>\n If A=2=>B=6 => C=0 but this is invalid as C is positive.<\/p>\n Similarly if A>2 C will be negative and cases are not valid.<\/p>\n Hence, A+B=6.<\/p>\n Question 3:\u00a0<\/b>Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick’s age is 1 year\u00a0less than the average age of all three, then Harry’s age, in years, is<\/p>\n 3)\u00a0Answer:\u00a018<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let tom’s age = x<\/p>\n =>\u00a0Dick=3x<\/p>\n =>harry = 6x<\/p>\n Given,<\/p>\n 3x+1 = (x+3x+6x)\/3<\/p>\n => x= 3<\/p>\n Hence, Harry’s age = 18 years<\/p>\n Question 4:\u00a0<\/b>Let k be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution if\u00a0and only if<\/p>\n a)\u00a0$\\mid k\\mid\\neq2$<\/p>\n b)\u00a0$\\mid k\\mid=2$<\/p>\n c)\u00a0$k\\neq2$<\/p>\n d)\u00a0$k=2$<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Two linear equations ax+by= c and dx+ ey = f have a unique solution if\u00a0$\\frac{a}{d}\\ne\\ \\frac{b}{e}$<\/p>\n Therefore, $\\frac{k}{4}\\ne\\ \\frac{1}{k}$ =>\u00a0$k^2\\ne\\ 4$<\/p>\n => k\u00a0$\\ne\\ $ |2|<\/p>\n Question 5:\u00a0<\/b>In May, John bought the same amount of rice and the same amount of wheat as he had\u00a0bought in April, but spent \u20b9 150 more due to price increase of rice and wheat by 20%\u00a0and 12%, respectively. If John had spent \u20b9 450 on rice in April, then how much did he\u00a0spend on wheat in May?<\/p>\n a)\u00a0Rs.560<\/p>\n b)\u00a0Rs.570<\/p>\n c)\u00a0Rs.590<\/p>\n d)\u00a0Rs.580<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let John buy “m” kg of rice and “p” kg of wheat.<\/p>\n Now let the price of rice be “r” in April. Price in May will be “1.2(r)”<\/p>\n Now let the price of wheat be “w” in April . Price in April will be “1.12(w)”.<\/p>\n Now he spent \u20b9150 more in May , so 0.2(rm)+0.12(wp)=150<\/p>\n Its also given that he had spent \u20b9450 on rice in April. So (rm)=450<\/p>\n So 0.2(rm)= (0.2)(450)=90 Substituting we get (wp)=60\/0.12 or (wp)=500<\/p>\n Amount spent on wheat in May will be 1.12(500)=\u20b9560<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>If x and y are non-negative integers such that $x + 9 = z$, $y + 1 = z$ and $x + y < z + 5$, then the maximum possible value of $2x + y$ equals<\/p>\n 6)\u00a0Answer:\u00a023<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We can write x=z-9 and y=z-1 Now we have x+y< z+5<\/p>\n Substituting we get z-9+z-1<z+5 or z<15<\/p>\n Hence the maximum possible value of z is 14<\/p>\n Maximum value of “x” is 5 and maximum value of “y” is 13<\/p>\n Now 2x+y = 10+13=23<\/p>\n Question 7:\u00a0<\/b>Aron bought some pencils and sharpeners. Spending the same amount of money as\u00a0Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one\u00a0sharpener is \u20b9 2 more than the cost of a pencil, then the minimum possible number of\u00a0pencils bought by Aron and Aditya together is<\/p>\n a)\u00a033<\/p>\n b)\u00a027<\/p>\n c)\u00a030<\/p>\n d)\u00a036<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of pencils bought by Aron\u00a0be “p” and the cost of each pencil be “a”.<\/p>\n Let the number of sharpeners bought Aron be “s” and the cost of each sharpener be “b”.<\/p>\n Now amount spent by Aron will be (pa)+(sb)<\/p>\n Aditya bought (2p) pencils and (s-10) sharpeners. Amount spent will be (2pa)+(s-10)b<\/p>\n Amount spent in both the cases is same<\/p>\n pa + sb = 2pa + (s-10)b or pa=10b<\/p>\n Now its given in the question that cost of sharpener is 2 more than pencil i.e. b=a+2<\/p>\n pa= 10a+20 or a=20\/(p-10)<\/p>\n Now the number of pencils has to be minimum, for that we have to find smallest “p” such that both “p” and “a” are integers. The smallest such value is p=11 . Total number of pencils bought will be p+2p=11+22=33<\/p>\n Question 8:\u00a0<\/b>A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is<\/p>\n a)\u00a011<\/p>\n b)\u00a013<\/p>\n c)\u00a010<\/p>\n d)\u00a012<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the cost of an apple, an orange and a mango be a, o, and m respectively.<\/p>\n Then it is given that:<\/p>\n 2a+4o+6m = a+4o+8m<\/p>\n or a = 2m.<\/p>\n Also, a+4o+8m = 8o + 7m<\/p>\n 10m-7m = 4o<\/p>\n 3m = 4o.<\/p>\n We can now express the cost of a basket in terms of mangoes only:<\/p>\n 2a+4o+6m = 4m+3m+6m = 13m.<\/p>\n Question 9:\u00a0<\/b>If $3x+2\\mid y\\mid+y=7$ and $x+\\mid x \\mid+3y=1$ then $x+2y$ is:<\/p>\n a)\u00a0$-\\frac{4}{3}$<\/p>\n b)\u00a0$\\frac{8}{3}$<\/p>\n c)\u00a0$0$<\/p>\n d)\u00a0$1$<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We need to check for all regions:<\/p>\n x >= 0, y >= 0<\/p>\n x >= 0, y < 0<\/p>\n x < 0, y >= 0<\/p>\n x < 0, y < 0<\/p>\n However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.<\/p>\n Let us start with\u00a0x >= 0, y >= 0,<\/strong><\/p>\n 3x + 3y = 7<\/p>\n 2x + 3y = 1<\/p>\n Hence, x = 6 and y = -11\/3<\/p>\n Since y > = 0, this is not satisfying the set of rules.<\/p>\n Next, let us test\u00a0x >= 0, y < 0,<\/strong><\/p>\n 3x – y = 7<\/p>\n 2x + 3y = 1<\/p>\n Hence, y = -1<\/p>\n x = 2.<\/p>\n This satisfies both the conditions. Hence, this is the correct point.<\/p>\n WE need the value of x + 2y<\/p>\n x + 2y = 2 + 2(-1) = 2 – 2 = 0.<\/p>\n Question 10:\u00a0<\/b>Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?<\/p>\n a)\u00a017<\/p>\n b)\u00a016<\/p>\n c)\u00a018<\/p>\n d)\u00a015<\/p>\n e)\u00a019<\/p>\n